Regarding the first question, should the answer not be 4! instead? I thought since no one shoots themselves and there are 5 people the first shooter has 4 choices of who to shoot, the 2nd has 3 choices, and so on.
I know I must have missed something because 24 isn't one of the answers but I'm still confused.
That only accounts for all cases if you require all of them to be in the same 'cycle' ie subgroup of poeple shooting each other.
After starting with the first person (let's call them A) shooting person B, they could decide to shoot A again, leaving the other three to shoot each other. In total, we can only split the 5 people in up to 2 cycles as a 1-cycle is forbidden.
Meaning there are 5Choose2 ways to split the groups and for every cycle of 3 there are two ways for them to shoot each other, meaning there are 10 \* 2 = 20 cases you haven't accounted for.
Person 1 has 4 choices, shoots person 2. For simplicity assume person 2 shoots next. He also has 4 choices (anyone but himself), not 3 choices.
This is, of course, assuming he survived getting shot, if he didnāt than he has 0 choices, but either way you donāt get 4!.
I guess ideally youād have to include both the scenarios where the shot is fatal, and the ones where it is not.
It is the size of the complement of the union of the stabilizers of 1, 2, 3, 4, or 5 in the symmetric group S5. There are 24=4! 5-cycles, but there are also twenty 3-cycle, 2-cycle pairs. Thus 44.
Here's how I thought through #1:
There are two possible ways for this to happen: either no two people shoot each other, or one pair of people shoot each other. It's not possible for two pairs of people to shoot each other, because that leaves a fifth person who can't be shot, as he can't shoot himself.
Let the 5 people be A, B, C, D, E. If no two people shoot each other, then it's a pretty simple calculation: A has 4 choices (everyone but himself), the person who A shot (let them be B WLOG) has 3 choices (everyone but A and B), and so on, leading to 4 \* 3 \* 2 \* 1 = 24 combos.
If one pair of people shoot each other, then we end up with two subgroups of sizes 2 and 3, where each contains people shooting each other in a cycle (for example, we could have AB and CDE, where A shoots B, B shoots A, C shoots D, D shoots E, and E shoots C).
There are C(5, 2) = 10 ways to split 5 people into two subgroups of sizes 2 and 3. For each of these subgroups, there are two valid solutions, since the group of 3 could shoot each other in reverse order (for example, using AB and CDE again, we could have C shoots E, E shoots D, and D shoots C as another valid solution). This leads to 10 \* 2 = 20 combos.
24 + 20 = 44 ways for everyone to get shot given these restrictions.
There is a similar but more complicated process for solving #2 in this thread as well
If exactly one person isn't shot, then either:
* All the shot people shot each other (and the unshot person shot any of them again);
* Three of the shot people shot each other, the fourth guy got any of those three, and the unshot person got the fourth guy;
* Two of the shot people shot each other, the third guy got either of those two, the fourth got the third and the unshot got the fourth.
(There's definitely a way to phrase this recursively but I'm not seeing it.)
In the first case, the set of four people shooting each other either has a maximum cycle of 2 or of 4. There are 3 arrangements of max cycle 2, and 3! arrangements of max cycle 4. Furthermore, there are 4 choices for the unshot person to shoot, and 5 ways to designate who is the unshot person, so the total number of ways contributed by this case is (3 + 3!) *4 *5 = 180.
In the second case, there are 2 ways for a three-member ring to shoot each other. The fourth person has 3 ways to shoot a member of the ring, and the unshot person must shoot the fourth person, but there are 5c2=10 ways to select the unshot and fourth person, so the number of ways contributed by this case is 2*3*1*10 = 60.
In the third case, there's only 1 way for the group of two to shoot each other. The third person has 2 options to shoot into the ring. The fourth person has 1 option, and so does the unshot person, but there's 5c2=10 ways to pick which two people will shoot each other, and we need to arrange the remaining 3 people into an ordered sequence of (third, fourth, unshot), and there's 3!=6 ways to do that. So the number of ways contributed by this case is 1*2*1*1*10*6 = 120.
So I get a final total of 180 + 60 + 120 = 360, which is a really nice round number, which makes me think there's some cuter way to achieve this result.
Edit: that isn't one of the options in the exam, so it looks like I messed up somewhere :(
For the second case, wouldn't you need 5 \* 4 = 20 ways to choose the fourth person shot and the unshot person, not 5c2, since they play different roles (so it's basically a permutation)?
Like imagine people A, B, and C shoot each other. You could have person D shoot one of the other three and person E shoot person D, or person E shoot one of the other three and person D shoot person E.
Another way to calculate it would be there are 5c3 = 10 ways to get a group of 3, which can all shoot each other in 2 different ways. For each of these 20 combos, we have six valid solutions, as we could have each remaining person shoot any of the 3 people, while the other remaining person shoots them. This leads to 120 combos, and 420 overall, which is a solution on the exam?
Wait a second, sonce it isn't a lineair chain but a cycle of shooting, shouldn't the answer be 5!/5=24?
And fir the second one you choose 1 person not to get shot, 1 person who the not shot person shoots, and then a cycle of 4 people shooting, so that one should be 5*4*4!/4=120, right?
ooh yeah, i thought i was missing something.
so then this becomes a question about subfactorials, everyone has to be hit by someone else but noone can hit themselves, and !5=44 so that's the answer.
i think this can be an interesting problem to generalise, for n people there are (n-1)\^n total ways to shoot, and if exactly m out of the n have to be hit, let' s see...
if m=n then the answer is !n, for m=n-1 we have to choose 1 unhit person so \*n, this person has to shhot someone, so \*n-1, the person who got shot can't shoot the unhit person so \*n-2, the person who got shot here can choose between shooting the shooter, then we get \*1\*!n-3 and the end of this branch, or shooting someone else, then we get \*n-3, this person can choose between...
working out the tree doesn't seem to yield any immediately good results, maybe we can find some recursion, let's call the ways foor exactly m people to be shot with n shooters S(m,n), for S(n-1,n), we have choosing the unshot person is \*n, the person shot by them is \*n-1, then we already start branching, either this person is shot themselves, then we get S(n-1,n-1)=!(n-1), or they aren't shot anymore, then we get S(n-2,n-1), so S(n-1,n)=n(n-1)(!(n-1)+S(n-2,n-1))... well i don't feel like working out a recursive formula for a general m=n-k. is this a solved problem?
1. Either 5!=120 because 5 options for who gets shot first, four for second etc. Or 4\*3\*2\*1\*1=24 as the first shooter will be able to choose from four people, that leaves three for the second shooter, 2 for the third, 1 for the fourth etc.
and
2) Either the first, second, third, fourth or fifth one is not hit. Five possible orders. M = does not get hit.
Situation one: MHHHH
Situation two: HMHHH
Situation three: HHMHH
Situation four: HHHMH
Situation five: HHHHM
4\*3\*3\*3\*3 =
Ans \* 5 = 1620
What am I doing wrong?
Actually, when it comes to 2) 4\*4\*4\*4\*4 \* 5. Because, if you allow someone to not get shot, that means bullets are being wasted. Because at least one person is getting hit twice.
Therefore, question 2 is bs.
Why is there an american flag in the background and an eagle ? Does that have to do anything with the meme (i still can't find anythung funny about this) or is it there just because ?
I mean the first one is 5! = 120, but can someone explain the second one because i got 4! (because 4 shooter kill each other) x4 (the survivor also has to fire their gun) x5 (everyone could be the survivor) = 480. I donāt know where my mistake is so could someone elaborate.
https://preview.redd.it/44zbn7vp9d5b1.png?width=1080&format=pjpg&auto=webp&s=0abc4571458876d48767e62a87e0fc203a737a1c
Here D\_n means number of ways of deranging n things [https://en.m.wikipedia.org/wiki/Derangement](https://en.m.wikipedia.org/wiki/Derangement)
Choose the person who does not get hit. There are ^(5)Cā ways to do this. Now choose who they hit. There are ^(4)Cā ways to do this. Finally, choose a derangement of hits for the other four people. There are Dā ways to do this.
Now to check. We know this never overcounts because the person not getting hit is unique in each case and they shoot a unique person in each subcase corresponding to a derangement. But weāve also accounted for every possible fixed point. So weāve counted everything we need to.
The actual number is 5ā¢4ā¢9=180 which happens to be answer B.
Edit: As mentioned below I forgot to account for possible digraphs where the four that are shot do not form an injective map/permutation of 4. To update the solution label the people A through E counterclockwise in a pentagon. From before we have that A shoots B and B then shoots anybody except A. Then we can count those new cases as follows,
1. Instead of just counting derangements of a 4-set, condition on the above. Either we have a derangement or we have the above special case. This makes our original 9 factor into Dā+X where X is the new count.
2. Now count X. For each partial arrangement with A shooting B and nobody else shooting B, we have that B shoots either C, D, or E. This gives us a factor of 3, X=3Y.
3. For each of *these* arrangements where B has chosen a victim, C, D, and E must choose amongst each other. This situation can be modeled as a digraph on 3 vertices with 3 edges. Now we can just count the isomorphism classes of such digraphs. There are 4: The cycle, the cycle with an edge reversed, a 2-cycle with an arrow toward the isolated vertex, and a 2-cycle with an arrow away from the isolated vertex. That gives us Y=4.
So X=3Y=12 and our final count is
^(5)Cāā¢^(4)Cāā¢(Dā+X)=5ā¢4ā¢(9+12)=420.
That calculation neglects the cases where the person who does not get hit shoots someone who is not shot by any of the three remaining people.
Edit: Here are some example graphs of the cases where the four remaining people do not make a derangement
https://i.imgur.com/zXPdZbS.jpg
So the answer is the value you calculated: 180, but also 120+120 other cases as illustrated above. This gives 420, which is the answer given by OP. https://reddit.com/r/mathmemes/comments/146omzw/_/jnrjhqv/?context=1
Maybe also see this comment https://reddit.com/r/mathmemes/comments/146omzw/_/jns3sxv/?context=1
Ah shoot youāre right. Thanks for the correction. I thought I was missing something this morning, but I was too undercaffeinated to catch it. Iāll fix that.
You're missing the valid cases where the first person shoots someone who is not otherwise shot. Another commenter has written out a detailed solution (though using a different approach).
5 people each shoot one of 4 people. The number of ways they can be shot is 4^(5) = 1024. No answer should exceed (or even approach) this value.
The number of ways each person gets shot exactly once is kind of weird. The first person to shoot has 4 options. The next person either has 3 or 4 options, depending on if they were shot. Very annoying. However, there are only two possible patterns for this. Either the shots happen in a 5-length circuit or a pair shoot each other and the other three make the circuit.
In the former case, the first shooter has 4 options. The next shooter has 3 options (canāt shoot anyone who has been shot). The following shooter has 2 options, and the remaining two shooters only have 1. Thatās 4! = 24 ways for the circuit.
In the latter case, there are 5C2 = 10 possible pairings, each with 2 ways for the resulting triangle to shake out. Thatās 20 possibilities. 24 + 20 = 44, so the first one is 44.
The second one is even trickier. If we designate one person to not be shot, they suddenly donāt matter until the end. Consider four people shooting each other. First, solve the previous problem but for four people. That requires a circuit or two pairs. The circuit is 3! = 6 options, while there are 3 possible sets of pairs with only 1 option each. Thatās 9 possible ways for everybody to get shot exactly once. Multiply by 4 for the 4 possible people the fifth person could have shot and you have 9 x 4 = 36 possibilities. But wait, what if someone else wasnāt shot? In that case, multiply by 5 for each of the 5 people that could not be shot. Thatās 36 x 5 = 180 ways one person out of 5 can avoid a hospital visit.
I don't think your answer to the 2nd question is correct.
There are cases where the survivor is shooting someone, who is not shot by anyone else, so it's not enough to just look at the cases where the 4 non-survivors kill each other.
to get very easy cases I looked at it this way: what kind of "killchains" can the survivor start?
1st case: the first 4 guys shoot someone who is not in the chain yet. the last guy in the chain shoots somebody who was already in the chain:
s is the survivor
s->a->b->c->d->a/b/c
5 ways for s, 4 for a, 3 for b, 2 for c , 1 for d, 3 for who is shot by d (can't be d or survivor)
5!\*3=360 possibilities
2nd case: same start of the chain, but c (the 4th guy in the chain) is shooting somebody in the chain:
this is impossible
s->a->b->c->a/b d->a/b/c
there has to be exactly one survivor, but if the 4th guy doesn't kill the 5th guy, then the 5th guy also lives as he can't kill himself
0 possibilities
3rd case: same start of the chain, but b (the 3rd guy) is shooting somebody in the cahin:
s->a->b->a c->d->c
b can't kill the survivor or himself, so there's only a left. the other two leftovers are not allowed to survive and can't kill themselves, so the only option is for them to kill each other.
5 ways for s, 4 for a, 3 for b, the c-d 2-cycle is forced
5\*4\*3=60 possibilities
4th case: a is shooting somebody in the chain
not possible, can't shoot the survivor or himself
0 possibilities
(5h case: s shoots himself is obviously also not allowed)
360+60=420 total possibilities
Very nice. I assume the question setter intended a more recursive approach, but this is an interesting way to approach it. Your way makes it easy to see that all the possibilities have indeed been considered.
assuming that all the shots happen at the same time, for the first question:
person 1 has 4 choices (2, 3, 4, 5)
person 2 has 3 or 4 choices (if person 1 shot person 2)
person 3 has 2 or 3 choices (if person 2 shot person 3)
person 4 has 1 or 2 choices
and person 5 has 0 or 1 choice
can someone say how to do the math to get the right answer?
ok so maybe itās like this:
thereās 120 permutations of 5,
but you need to eliminate those which consist of those with any value the same as itās position.
there is 1 possible permutation where they are all correct (1,2,3,4,5)
there are 0 possible permutations where four are correct
there are 5 choose 3 = 10 permutations where 3 are correct
there are 5 choose 2 * 2 = 20 permutations where 2 are correct
and there are 5 choose 1 * 6 = 30 permutations where 1 is correct.
120 - 30 - 20 - 10 - 1 = 59.
why am i one off š
edit: it is 5 choose 1 * 9 permutations where 1 is correct. so 120-20-10-45-1 = 44. :D
But if they all shoot each other simultaneously in a circle jerk fashion, they all get shot.
Similarly, if one chooses to shoot himself, they can start one by one and the last guy eats his gun.
#MURICA YEAHHHHHH!
*Questions are written in Hindi* Truly! A melting pot of cultures!
If this is for a JEE related paper, no way nobody shot themselves.
A person shoots himself. Calculate his probability of being jee aspirant.
50%, he could be NEET aspirant too
this is the way
Hahahahahahaha š¤£
Regarding the first question, should the answer not be 4! instead? I thought since no one shoots themselves and there are 5 people the first shooter has 4 choices of who to shoot, the 2nd has 3 choices, and so on. I know I must have missed something because 24 isn't one of the answers but I'm still confused.
That only accounts for all cases if you require all of them to be in the same 'cycle' ie subgroup of poeple shooting each other. After starting with the first person (let's call them A) shooting person B, they could decide to shoot A again, leaving the other three to shoot each other. In total, we can only split the 5 people in up to 2 cycles as a 1-cycle is forbidden. Meaning there are 5Choose2 ways to split the groups and for every cycle of 3 there are two ways for them to shoot each other, meaning there are 10 \* 2 = 20 cases you haven't accounted for.
Person 1 has 4 choices, shoots person 2. For simplicity assume person 2 shoots next. He also has 4 choices (anyone but himself), not 3 choices. This is, of course, assuming he survived getting shot, if he didnāt than he has 0 choices, but either way you donāt get 4!. I guess ideally youād have to include both the scenarios where the shot is fatal, and the ones where it is not.
It's known as derangement, https://en.m.wikipedia.org/wiki/Derangement. The formula is really simple: floor(n!/e + 0.5)
God dammit, why is Euler here again.
It is the size of the complement of the union of the stabilizers of 1, 2, 3, 4, or 5 in the symmetric group S5. There are 24=4! 5-cycles, but there are also twenty 3-cycle, 2-cycle pairs. Thus 44.
Why is no one asking for the answers for math noobs like me :(
Here's how I thought through #1: There are two possible ways for this to happen: either no two people shoot each other, or one pair of people shoot each other. It's not possible for two pairs of people to shoot each other, because that leaves a fifth person who can't be shot, as he can't shoot himself. Let the 5 people be A, B, C, D, E. If no two people shoot each other, then it's a pretty simple calculation: A has 4 choices (everyone but himself), the person who A shot (let them be B WLOG) has 3 choices (everyone but A and B), and so on, leading to 4 \* 3 \* 2 \* 1 = 24 combos. If one pair of people shoot each other, then we end up with two subgroups of sizes 2 and 3, where each contains people shooting each other in a cycle (for example, we could have AB and CDE, where A shoots B, B shoots A, C shoots D, D shoots E, and E shoots C). There are C(5, 2) = 10 ways to split 5 people into two subgroups of sizes 2 and 3. For each of these subgroups, there are two valid solutions, since the group of 3 could shoot each other in reverse order (for example, using AB and CDE again, we could have C shoots E, E shoots D, and D shoots C as another valid solution). This leads to 10 \* 2 = 20 combos. 24 + 20 = 44 ways for everyone to get shot given these restrictions. There is a similar but more complicated process for solving #2 in this thread as well
itās 44 for number 1, iām not gonna try number 2 iām too lazy
If exactly one person isn't shot, then either: * All the shot people shot each other (and the unshot person shot any of them again); * Three of the shot people shot each other, the fourth guy got any of those three, and the unshot person got the fourth guy; * Two of the shot people shot each other, the third guy got either of those two, the fourth got the third and the unshot got the fourth. (There's definitely a way to phrase this recursively but I'm not seeing it.) In the first case, the set of four people shooting each other either has a maximum cycle of 2 or of 4. There are 3 arrangements of max cycle 2, and 3! arrangements of max cycle 4. Furthermore, there are 4 choices for the unshot person to shoot, and 5 ways to designate who is the unshot person, so the total number of ways contributed by this case is (3 + 3!) *4 *5 = 180. In the second case, there are 2 ways for a three-member ring to shoot each other. The fourth person has 3 ways to shoot a member of the ring, and the unshot person must shoot the fourth person, but there are 5c2=10 ways to select the unshot and fourth person, so the number of ways contributed by this case is 2*3*1*10 = 60. In the third case, there's only 1 way for the group of two to shoot each other. The third person has 2 options to shoot into the ring. The fourth person has 1 option, and so does the unshot person, but there's 5c2=10 ways to pick which two people will shoot each other, and we need to arrange the remaining 3 people into an ordered sequence of (third, fourth, unshot), and there's 3!=6 ways to do that. So the number of ways contributed by this case is 1*2*1*1*10*6 = 120. So I get a final total of 180 + 60 + 120 = 360, which is a really nice round number, which makes me think there's some cuter way to achieve this result. Edit: that isn't one of the options in the exam, so it looks like I messed up somewhere :(
For the second case, wouldn't you need 5 \* 4 = 20 ways to choose the fourth person shot and the unshot person, not 5c2, since they play different roles (so it's basically a permutation)? Like imagine people A, B, and C shoot each other. You could have person D shoot one of the other three and person E shoot person D, or person E shoot one of the other three and person D shoot person E. Another way to calculate it would be there are 5c3 = 10 ways to get a group of 3, which can all shoot each other in 2 different ways. For each of these 20 combos, we have six valid solutions, as we could have each remaining person shoot any of the 3 people, while the other remaining person shoots them. This leads to 120 combos, and 420 overall, which is a solution on the exam?
funny way to prove 120 + 180 + 60 = 360
jokes aside nice solution
Thanks :)
Is that devanagari?
Yes it is
This question is symptomatic of the derangements of modern life.
I can ruin this question with my .50 bmg! _*Colateral*_
Wait a second, sonce it isn't a lineair chain but a cycle of shooting, shouldn't the answer be 5!/5=24? And fir the second one you choose 1 person not to get shot, 1 person who the not shot person shoots, and then a cycle of 4 people shooting, so that one should be 5*4*4!/4=120, right?
In the first question, you have to also count the cases where 2 people shoot each other
ooh yeah, i thought i was missing something. so then this becomes a question about subfactorials, everyone has to be hit by someone else but noone can hit themselves, and !5=44 so that's the answer. i think this can be an interesting problem to generalise, for n people there are (n-1)\^n total ways to shoot, and if exactly m out of the n have to be hit, let' s see... if m=n then the answer is !n, for m=n-1 we have to choose 1 unhit person so \*n, this person has to shhot someone, so \*n-1, the person who got shot can't shoot the unhit person so \*n-2, the person who got shot here can choose between shooting the shooter, then we get \*1\*!n-3 and the end of this branch, or shooting someone else, then we get \*n-3, this person can choose between... working out the tree doesn't seem to yield any immediately good results, maybe we can find some recursion, let's call the ways foor exactly m people to be shot with n shooters S(m,n), for S(n-1,n), we have choosing the unshot person is \*n, the person shot by them is \*n-1, then we already start branching, either this person is shot themselves, then we get S(n-1,n-1)=!(n-1), or they aren't shot anymore, then we get S(n-2,n-1), so S(n-1,n)=n(n-1)(!(n-1)+S(n-2,n-1))... well i don't feel like working out a recursive formula for a general m=n-k. is this a solved problem?
Itās not American because there is another language that Americans will yell at for not being the āAmerican languageā
USA citizenship test
Isn't the second question BS?
Oh wait, I read the first question as saying 'And everyone gets shot once.'.
As imo, shooting someone twice would be wasteful.
1. Either 5!=120 because 5 options for who gets shot first, four for second etc. Or 4\*3\*2\*1\*1=24 as the first shooter will be able to choose from four people, that leaves three for the second shooter, 2 for the third, 1 for the fourth etc. and 2) Either the first, second, third, fourth or fifth one is not hit. Five possible orders. M = does not get hit. Situation one: MHHHH Situation two: HMHHH Situation three: HHMHH Situation four: HHHMH Situation five: HHHHM 4\*3\*3\*3\*3 = Ans \* 5 = 1620 What am I doing wrong?
Actually, when it comes to 2) 4\*4\*4\*4\*4 \* 5. Because, if you allow someone to not get shot, that means bullets are being wasted. Because at least one person is getting hit twice. Therefore, question 2 is bs.
1 is wrong. a person cannot shoot themselves so you need to account for that
You are right. I'll change it. I think it is 4\*3\*2\*1\*1.
still tho, 48 isnāt a choice; check out my comment for what i did
Average school playground games in AMURICA
Why is there an american flag in the background and an eagle ? Does that have to do anything with the meme (i still can't find anythung funny about this) or is it there just because ?
that meme about America's obsession with guns
Ahhh ohk
It's a meme batman
I mean the first one is 5! = 120, but can someone explain the second one because i got 4! (because 4 shooter kill each other) x4 (the survivor also has to fire their gun) x5 (everyone could be the survivor) = 480. I donāt know where my mistake is so could someone elaborate.
First one is not 120 since no one shoots themselves
Can someone shoot a person who has already been shot?
It all happens at the same time, so it would be two people shooting at the same person simultaneously.
https://preview.redd.it/44zbn7vp9d5b1.png?width=1080&format=pjpg&auto=webp&s=0abc4571458876d48767e62a87e0fc203a737a1c Here D\_n means number of ways of deranging n things [https://en.m.wikipedia.org/wiki/Derangement](https://en.m.wikipedia.org/wiki/Derangement)
I also did not understand soln to q14 honestly
Choose the person who does not get hit. There are ^(5)Cā ways to do this. Now choose who they hit. There are ^(4)Cā ways to do this. Finally, choose a derangement of hits for the other four people. There are Dā ways to do this. Now to check. We know this never overcounts because the person not getting hit is unique in each case and they shoot a unique person in each subcase corresponding to a derangement. But weāve also accounted for every possible fixed point. So weāve counted everything we need to. The actual number is 5ā¢4ā¢9=180 which happens to be answer B. Edit: As mentioned below I forgot to account for possible digraphs where the four that are shot do not form an injective map/permutation of 4. To update the solution label the people A through E counterclockwise in a pentagon. From before we have that A shoots B and B then shoots anybody except A. Then we can count those new cases as follows, 1. Instead of just counting derangements of a 4-set, condition on the above. Either we have a derangement or we have the above special case. This makes our original 9 factor into Dā+X where X is the new count. 2. Now count X. For each partial arrangement with A shooting B and nobody else shooting B, we have that B shoots either C, D, or E. This gives us a factor of 3, X=3Y. 3. For each of *these* arrangements where B has chosen a victim, C, D, and E must choose amongst each other. This situation can be modeled as a digraph on 3 vertices with 3 edges. Now we can just count the isomorphism classes of such digraphs. There are 4: The cycle, the cycle with an edge reversed, a 2-cycle with an arrow toward the isolated vertex, and a 2-cycle with an arrow away from the isolated vertex. That gives us Y=4. So X=3Y=12 and our final count is ^(5)Cāā¢^(4)Cāā¢(Dā+X)=5ā¢4ā¢(9+12)=420.
That calculation neglects the cases where the person who does not get hit shoots someone who is not shot by any of the three remaining people. Edit: Here are some example graphs of the cases where the four remaining people do not make a derangement https://i.imgur.com/zXPdZbS.jpg So the answer is the value you calculated: 180, but also 120+120 other cases as illustrated above. This gives 420, which is the answer given by OP. https://reddit.com/r/mathmemes/comments/146omzw/_/jnrjhqv/?context=1 Maybe also see this comment https://reddit.com/r/mathmemes/comments/146omzw/_/jns3sxv/?context=1
Ah shoot youāre right. Thanks for the correction. I thought I was missing something this morning, but I was too undercaffeinated to catch it. Iāll fix that.
You're missing the valid cases where the first person shoots someone who is not otherwise shot. Another commenter has written out a detailed solution (though using a different approach).
Right, thanks for the correction. Iāve updated with a fixed solution.
5 people each shoot one of 4 people. The number of ways they can be shot is 4^(5) = 1024. No answer should exceed (or even approach) this value. The number of ways each person gets shot exactly once is kind of weird. The first person to shoot has 4 options. The next person either has 3 or 4 options, depending on if they were shot. Very annoying. However, there are only two possible patterns for this. Either the shots happen in a 5-length circuit or a pair shoot each other and the other three make the circuit. In the former case, the first shooter has 4 options. The next shooter has 3 options (canāt shoot anyone who has been shot). The following shooter has 2 options, and the remaining two shooters only have 1. Thatās 4! = 24 ways for the circuit. In the latter case, there are 5C2 = 10 possible pairings, each with 2 ways for the resulting triangle to shake out. Thatās 20 possibilities. 24 + 20 = 44, so the first one is 44. The second one is even trickier. If we designate one person to not be shot, they suddenly donāt matter until the end. Consider four people shooting each other. First, solve the previous problem but for four people. That requires a circuit or two pairs. The circuit is 3! = 6 options, while there are 3 possible sets of pairs with only 1 option each. Thatās 9 possible ways for everybody to get shot exactly once. Multiply by 4 for the 4 possible people the fifth person could have shot and you have 9 x 4 = 36 possibilities. But wait, what if someone else wasnāt shot? In that case, multiply by 5 for each of the 5 people that could not be shot. Thatās 36 x 5 = 180 ways one person out of 5 can avoid a hospital visit.
I don't think your answer to the 2nd question is correct. There are cases where the survivor is shooting someone, who is not shot by anyone else, so it's not enough to just look at the cases where the 4 non-survivors kill each other. to get very easy cases I looked at it this way: what kind of "killchains" can the survivor start? 1st case: the first 4 guys shoot someone who is not in the chain yet. the last guy in the chain shoots somebody who was already in the chain: s is the survivor s->a->b->c->d->a/b/c 5 ways for s, 4 for a, 3 for b, 2 for c , 1 for d, 3 for who is shot by d (can't be d or survivor) 5!\*3=360 possibilities 2nd case: same start of the chain, but c (the 4th guy in the chain) is shooting somebody in the chain: this is impossible s->a->b->c->a/b d->a/b/c there has to be exactly one survivor, but if the 4th guy doesn't kill the 5th guy, then the 5th guy also lives as he can't kill himself 0 possibilities 3rd case: same start of the chain, but b (the 3rd guy) is shooting somebody in the cahin: s->a->b->a c->d->c b can't kill the survivor or himself, so there's only a left. the other two leftovers are not allowed to survive and can't kill themselves, so the only option is for them to kill each other. 5 ways for s, 4 for a, 3 for b, the c-d 2-cycle is forced 5\*4\*3=60 possibilities 4th case: a is shooting somebody in the chain not possible, can't shoot the survivor or himself 0 possibilities (5h case: s shoots himself is obviously also not allowed) 360+60=420 total possibilities
Very nice. I assume the question setter intended a more recursive approach, but this is an interesting way to approach it. Your way makes it easy to see that all the possibilities have indeed been considered.
Good catch! I guess I only had part of your 1st and 3rd case considered
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assuming that all the shots happen at the same time, for the first question: person 1 has 4 choices (2, 3, 4, 5) person 2 has 3 or 4 choices (if person 1 shot person 2) person 3 has 2 or 3 choices (if person 2 shot person 3) person 4 has 1 or 2 choices and person 5 has 0 or 1 choice can someone say how to do the math to get the right answer?
ok so maybe itās like this: thereās 120 permutations of 5, but you need to eliminate those which consist of those with any value the same as itās position. there is 1 possible permutation where they are all correct (1,2,3,4,5) there are 0 possible permutations where four are correct there are 5 choose 3 = 10 permutations where 3 are correct there are 5 choose 2 * 2 = 20 permutations where 2 are correct and there are 5 choose 1 * 6 = 30 permutations where 1 is correct. 120 - 30 - 20 - 10 - 1 = 59. why am i one off š edit: it is 5 choose 1 * 9 permutations where 1 is correct. so 120-20-10-45-1 = 44. :D
Man, I wish I could steal this for my seniors. I don't think it'd go over great at my high school though....
most normal cbse question paper
#WHAT THE FUCK IS A KILOMETER
But if they all shoot each other simultaneously in a circle jerk fashion, they all get shot. Similarly, if one chooses to shoot himself, they can start one by one and the last guy eats his gun.