There are 4 boys & 6 prizes can be distributed among them such that each has at least 1 prize.How many ways that can be done?

Please give a detailed approach.

& yes, If the cases are : 1) 3 1 1 1

                                       2) 2 2 1 1   ----- prizes given to boys.

Then what will be the expected output
asked May 25 in Pre-Algebra Answers by Subarna Das (440 points)

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This question can be taken to different levels, with increasing detail.

The first level is the distribution combination. Each boy must have one prize so that leaves two more prizes to be distributed. So one boy can be given both so that one boy has three prizes and the others have one each. That’s Combination 1.

Then we can give two of the boys an extra prize each, so two of the boys have two prizes each and the other two boys have only one each. That’s Combination 2.

The next level involves permutations. To make it more personal and more interesting, we can name the boys: Ali, Baba, Chaz and Dave.

For Combination 1, there are 4 permutations: the lucky boy with 3 prizes can be any one of the four. So if Ali is the lucky one, that’s C1P1; if it’s Baba, that’s C1P2; if Chaz, it’s C1P3; and if Dave, it’s C1P4.

For Combination 2, there are 6 perms: Ali and Baba get 2 each, Chaz and Dave get one each, that’s C2P1; Ali and Chaz get 2, Baba and Dave get one, that’s C2P2; Ali and Dave get 2, Baba and Chaz get 1, that’s C2P3; Baba and Chaz get 2, Ali and Dave get 1, that’s C2P4; Baba and Dave get 2, Ali and Chaz get 1, that’s C2P5; Chaz and Dave get 2, Ali and Baba get 1, that’s C2P6.

At level 2 there are 10 ways the prizes can be distributed.

At level 3 we have 4 distinct boys and 6 distinct prizes.

Consider Combination 1. First, select the group of three prizes. We have a choice of 6 prizes for the first one of the three prizes, a choice of one of 5 prizes for the next of the three prizes, and a choice of 4 for the last. That would make 6×5×4 ways of choosing three, but these include all ways of ordering the prizes. We are only interested in the combination, not the order in which they were picked. So it’s 6×5×4÷6=20 combinations, because there are 6 ways of arranging 3 objects in order. The remaining 3 prizes can be arranged in 6 ways.

So for every combination of the trio of prizes we have 6 ways of arranging the remaining 3 single prizes. That makes 120 ways in all.

Next, Combination 2. There are 6×5÷2=15 distinct ways to pick two prizes out of 6 and there are 4×3÷2=6 ways to pick another two prizes out of the remaining 4. The remaining two prizes can be arranged in two ways. In total there are 15×6×2=180 ways.

Now we come to allocating prizes to the boys. First, we line them up in order ABCD. Let’s take the case where A (Ali) receives 3 prizes while the others get one each. So Ali could receive 1 of 20 sets of 3 prizes, and the others can be given prizes in 6 different ways. That’s 120 ways for that single combination, and there are 4 distinct combinations, depending on which boy gets 3 prizes. So there are 480 ways of distributing prizes for Combination 1.

For Combination 2 we identified C2P1-C2P6 as the 6 permutations of boys: 2-2-1-1.

Let’s take an example: Ali gets 1 prize, Baba gets 2 prizes, Chaz gets one, Dave gets 2. The two prizes Baba gets are 1 pair out of 15 possible pairs, and the two Dave gets are 1 pair out of 6 possible remaining pairs. Between them their are 90 ways of being allocated 4 prizes. The remaining two boys have 2 prizes between them so there are two ways they can be allocated. So in all we have 180 arrangements, and these apply to every permutation C2P1-C2P6, making 6×180=1080 in all.

Add the totals for each combination and we get 480+1080=1560 distinct arrangements for allocating 6 different prizes to 4 boys.

answered May 26 by Rod Top Rated User (545,980 points)
selected May 27 by Subarna Das
What happens when the prizes are identical (indistinguishable) ?

You simply stop at level 2, because level 3 is all about non-identical (distinct) prizes. So at level 2 the answer is 10, 4 for Combination 1 and 6 for Combination 2.

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