In how many ways can 5 different candiesbe distributed among

4 word answer : "No pain, no gain" --- something that my physical trainer taught me a few years ago.

Why do you have to divide by 2! on 2-2-1-N but don't have to divide by 2! on 3-1-1-N?

In the case of 2-2-1-N we divide by 2! because we choose 2 groups of 2 one after the other and order when choosing them doesn't matter. Anyway, we permute them latter (see the factor of 4!).
Similarly, in the case of 3-1-1-N we should divide by 2! if we choose one single ball after the other 4! * 5C3 * 2C1 * 1C1/2!.
If it is clear that we split the two remaining balls after choosing 3, we don't need the 2C1 and 1C1 factors. That's why 4! * 5C3.

Thanks for your reply Eva, but I still haven’t fully understood it yet.
Let’s use the 3-1-1-N example, let’s say the candies are named A, B, C, D and E:
One possible outcome would be ABC-D-E-N, for this outcome there are 4! possible ways to distribute this outcome among the four people and 5C3 ways for the one person to get 3 candies so there are 240 possible arrangements of this kind.
Now I don’t understand exactly how this would look like if we choose one candy after the other, so that this formula applies: 4! * 5C3 * 2C1 * 1C1/2! If you draw again ABC for the person who gets 3, you have D and E left for the next person so two choices and then you have again two choices about who out of the remaining two people gets the last candy?! So after ABC was picked you have 2C1*2!=4 options and you could permute this outcome in 4! ways, which obviously is wrong. How exactly does the process look like if I draw according to 4! * 5C3 * 2C1 * 1C1/2!?

How exactly does the process look like if I draw according to 4! * 5C3 * 2C1 * 1C1/2!?

Think of carrying out the process in two steps:
1) Divide the 5 candies into 3 groups - 3, 1, 1
Decide on the 3 candies somebody will get them - 5C3 - for example A, B, C (but think of all the possibilities)
Which one is the first 1 single candy - 2C1 - D or E
Which one is the second 1 single candy - 1C1 - E or D
This would give 5C3 * 2C1 * 1C1
Divide by 2! because we are going to permute the groups of 3, 1, 1, 0, so we don't care about which single candy was chosen first.
ABC, D, E is the same as ABC, E, D - what matters is that A, B, C are together, and D and E are single
2) After we have our groups of 3, 1, 1, 0 candies, we consider them as 4 distinct objects and just permute them - 4! - because we have 4 different kids.

I am getting 8!/(3!*5!)=56. I am using the stars bars method. (n+k-1)C(k-1). Why is this answer incorrect?

There are various methods of solving P&C questions and there are many formulas. You need to know the exact situation in which you can use a particular formula or better yet, you should understand the reason a particular method works in a particular situation.

Stars bars method (or the formula (n+k-1)C(k-1)) is useful only when the objects to be distributed are identical. You split identical objects into different groups. You divide by 5! because the objects are identical. Here you can not use that method because the candies are different.

Check out these posts on when to use which method:

https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/12 ... 93-part-1/
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/12 ... s-part-ii/

Certainly not. You will not have the same answer in every case.
You can use the same formula but the value of n will vary in each case.

Say if every kid must receive at least one mango, you take any 4 mangoes from the bunch of 10 (they are identical) and give one to each of the 4 kids. You cannot do this in more than 1 way because all the mangoes are the same and all the kids are receiving exactly one mango.
Next, you are left with 6 mangoes and 4 kids and you need to distribute these 6 among the kids such that a kid may get no mango and another may get all etc. (You can do this because all the kids have already got a mango each and hence our condition is already satisfied.)
This boils down to our previous question except that the value of n = 6 now, not n = 10.

Now you use the formula as (6 + 4 - 1)C(4 - 1) = 9!/6!*3!

You can do the same thing for at least 2 mangoes too.

Why do we ignore the case in which no candy is given to any child? It is literally in the question statement that the children could get zero candies:

Admittedly, I'm not a native English speaker but to me it's clear that I do not have to give a candy to any child. I can keep all candies to myself.

Nor does the question state that all five candies must be given out to children. I could give out 0, 1, 2, 3, 4, or all five.

So, using the awesome approach posted by shrikar23, we would get 5^5 because there are 5 ways of giving out each distinct candy: give it to no child, or to child 1, or to child 2 and so forth.

The question states that you need to "distribute 5 candies". So all 5 candies must be given out. It is possible that one or more children may not get any candy and one or more may get more than one candies but all 5 have to be distributed.

what will be answer if candies are similar?

I think there are some mistakes in the previous explications.
There are two ways of solving it:

- Fastest way
Instead of thinking you have to distribute candies among children you can think in distribute children among candies.
So you have 5 candies and 4 children. And you can "repeat" children. So it is a permutation with repetition of 4 children in 5 ways (candies): 4^5 = 1024

- Slowest way
5-N-N-N : 4!/3! * 5C5 = 4
4-1-N-N : 4!/2! * 5C4*1C1 = 12*5*1 = 60
3-2-N-N : 4!/2! * 5C3*2C2 = 12*10*1 = 120
3-1-1-N : 4!/2! * 5C3*2C1*1C1 = 12*10*2*1 = 240
2-2-1-N : 4!/2! * 5C2*3C2*1C1 = 12*10*3*1 = 360
2-1-1-1 : 4!/3! * 5C2*3C1*2C1*1C1 = 4*10*3*2*1 = 240

4+60+120+240+360+240 = 1024 = 4^5

Explication:
For example, in the 3-1-1-N row, 4!/2! are the ways you can arrange 3-1-1-N (1-1-3-N, 1-1-N-3, 1-3-1-N, 1-3-N-1, 1-N-1-3, 1-N-3-1, 3-1-1-N, 3-1-N-1, 3-N-1-1, N-1-1-3, N-1-3-1, N-3-1-1): total 12 ways. So you have decided the number of candies that each boy is going to receive.
But candies are different (as the statement says) so we have to choose wich candies you give to each child (according with the number chosen before). In this case order does not mind because if a child receives gummi bears, cotton candy and M&M is the same as he receives M&M, cotton candy and gummi bears. So we use combinations:
5C3*2C1*1C1. First we choose which candies will be for the child who receives 3 (5C3). After that we have 2 remaining candies. We can choose which candy to give to one of the children who receives one (2C1). And the remaining candy is given to the other child (1C1).

Agreed. Great analysis. andreagonzalez2k

I work it out a bit backwards (I suppose) but get the same result going through the slow way. Ive always followed this way to keep the calculations and reasoning straight in my head (although I think it adds an extra step):

(1st) for each case I divide the different items into identical “stacks”

(2nd)then I account for any over counting if we have “stacks” of the same size

(3rd)then I arrange each of the “identical” stacks and shuffle them among the 4 different children, treating any child who gets 0 more than once as “identical elements”


Case 1: 5-0-0-0

(1st) no. of ways to choose candies

5 c 5 = 1

(2nd) we can arrange this 1 stack among the 4 children in:

4!/3! Ways (we need to remove the arrangements in which the children who receive 0 are over counted in 4!)

(5 c 5) * (4!/3!) = 1 * 4 = 4


Case 2: 4-1-0-0

(1st) no of ways to break chocolate into “identical stacks”

(5 c 4) * (4 c 1) = 5

And

(2nd) for each of the ways to break down the Items into non-distinct stacks, we can shuffle the stacks among the 4 children:

4!/2! = 12


5 * 12 = 60


Case 3: 3-2-0-0

(1st) choose and select the candies to break into identical “stacks”

(5 c 3) * (2 c 2) = 10

And

(2nd) for each of the 10 ways to break down into non-distinct stacks, we can shuffle around and arrange the identical stacks in:

4!/2! = 4 * 3 = 12


10 * 12 = 120


Case 4: 3-1-1-0

(1st) here is where the way I do it may differ a little (though we end up with the same calculation): 1st, we choose the ways to break down the different chocolates into stacks:

(5 c 3) * (2 c 1) * (1 c 1)

But, for the stacks that are of the SAME SIZE (1 and 1) we will be over counting each one through the selection method - right now, we are ignoring the ordering. So multiply the above result by (1 / 2!)

(5! / 3! 1! 1!) * (1 / 2!) = 10 ways

And

(2nd) for each of these 10 ways to break down the different chocolates into non-distinct stacks, we can then shuffle the stacks among the 4 children in;

4! = 24 ways


10 * 24 = 240


Case 5: 2-2-1-0

(1st) breaking down into identical stacks, but again we have two stacks of the same size (and will be over counting)

(5 c 2) * (3 c 2) * (1 c 1) * (1 / 2!) = 15 ways

And

(2nd) for each of the 15 ways to beak the Choc down into non-distinct stacks, these stacks can be shuffled around the 4 children in:

4! = 24 ways


15 * 24 = 360

Case 6: 2-1-1-1

(1st) (5 c 2) (3 c 1) (2 c 1) (1 c 1) * (1 / 3!) = 10

And

(2nd) 4! = 24


10 * 24 = 240

Adding up the cases:

4 + 60 + 120 + 240 + 360 + 240 =

1024

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The problem of choosing the slowest way is not to waste too much time...

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