Show that at least 2 children will receive the same number of cakes, if 27 identical cakes are to be distributed to 8 children.

Question

Question: There are 27 identical cakes to be distributed to 8 children.

a) How many ways can you do so with no further restrictions?

b) How many ways can you distribute the cakes such that each child receives at least 2 cakes?

c) Show that at least 2 children will receive the same number of cakes.

I understand how to do parts (a) and (b) by using the placeholders and separators method (I've included them anyways just in case it helps) but I'm uncertain what to do for part (c)? My instincts from what I've been learning so far tells me that it's something related to the Pigeonhole Principle but I'm not too sure.

Answer 1

Start by giving each child a different number of cakes. Let kid 1 get one cake, kid 2 get two cakes and so on. After kid 6, $21$ cakes have been given out and you do not have 7 more cakes. Hence the remaining two kids (7 and 8) will have to get cakes with the sum being 7 cakes. But since 1, 2, 3, 4, 5 and 6 cakes have already been given out, each of the two will the same number of cakes as someone before.

Answer 2

If each child receives a different number of cakes, then we must distribute at least $0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28$ cakes. However, we only have $27$ cakes. Therefore, at least two children must receive the same number of cakes.

Answer 3

Suppose the scenario where the 1st child recieves 0 cakes, 2nd child recieves 1 cake, 3rd child recieves 2 cakes, 4th recieves 3 cakes, 5th recieves 4 cakes, 6th recieves 5 cakes, 7th recieves 6 cakes. There is only one child left and 6 cakes left. Therefore regardless the number of cakes the last child recieves, it will be the same number as one of the other child. Hence, at least two children will recieve the same number of cakes