Confusing combination-permutations question

Question

In a shop there are five types of ice-creams available. A child buys six ice-creams. Is it true that the number of different ways the child can buy six ice creams is equal to the number of different ways of arranging 6 A's and 4 B's in a row?

I tried it as follows:

The child problem- There are five kinds of ice-creams, so the child can select six ice-creams in $5^6=15625$ ways.

A's and B's problem- There are 6 A's and 4 B's. We can place these 10 items in a row in 10!/6!4! ways, which is equal to 210.

But the actual answer is true. How can it be? Where am I wrong?

Answer 1

The answer provided by you is wrong. The count of $5^6$ selects and arranges 6 ice-creams in a order after selecting from 5 different ice-creams. But we want to get no of ways by which the child can buy 6 ice-creams and not to arrange them.

The correct approach can be as follows as the total no of ice creams of each type are unknown .

Let no of ice-creams bought of each type as $a,b,c,d,e$ .So, $$a+b+c+d+e=6$$ as boy needs 6 ice-creams. And thus answer is $10_{C_4}$ . Which is same as arranging 6A's and 4B's in a row.

Answer 2

Solve this using Beggars method :-

Let the 5 different types of ice creams be $x_1$, $x_2$, $x_3$, $x_4$, $x_5$.

Now we can choose each ice cream once, more than once or none times. But the sum of number of ice creams chosen should be equal to 6. Therefore

$$ x_1 + x_2 + x_3 + x_4 + x_5 = 6 $$

Therefore number of different ice-creams = $p$ = 5
Number of ice creams to be taken or distributed = $n$ = 6

Therefore number of total ways to distribute is $$\binom{n + p - 1}{p - 1} = \binom{10}{4} = \frac{10!}{4!6!}$$

Now number of ways to arrange 6 As and 4 Bs in a row = $\frac{10!}{4!6!}$ as total letters are 10 and 6 of one kind are similar and 4 of other kind are similar.

Therefore $\frac{10!}{4!6!} = \frac{10!}{4!6!}$
Hence proved ..