About squirrels and boxes

Question

There is a squirrel, 8 different nuts and 2 boxes. How many ways does the squirrel have to put all 8 nuts in the boxes? I started solving it like that: For every nut there are two conditions: to be in the first or in the second box. Moreover, the choice is independent from other nuts, so the amount of ways is $$2\cdot 2\cdots 2 = 2^8$$ I think there is another solution by using $\mathrm C_n^k$, especially because we know that $$\mathrm C^n_0+\mathrm C^n_1+\mathrm C^n_2+\ldots+\mathrm C_n^n = 2^n$$. Can you please explain how to apply this formula to the problem?

Answer 1

Sure, it suffices to select which nuts we put into box $1$.

There are $\binom{8}{0}$ ways to do it if we put $0$ nuts in box $1$, there are $\binom{8}{1}$ ways to do it if we put $1$ nut in box $1$ , there are $\binom{8}{2}$ ways to do it if we put $2$ nuts in box $1$ etc $\dots$

Answer 2

It should be clarified whether the nuts are distinguishable. Note that while your solution works for distinguishable nuts, if the squirrel can't tell nuts apart (i.e. nuts 1,2,3,5 in box 1 is the same as nuts 1,4,5,8 in box 1, or any other 4 nuts for that matter), then this is a classic example of the stars-and-bars, or ball-and-urn technique.

The number of ways to put 8 indistingushable nuts in 2 boxes is $\binom{2+8-1}{8} = \binom{9}{8} = 9$ ways.

Answer 3

Yeap.   Summing over ways to select which from the eight distinct nuts go into the first box, equals counting ways to assign one from two boxes to each of the eight nuts.   This is the combinatorial proof of the equality.

$$\sum_{k=0}^8 {^8\mathrm C_k} = 2^8$$

${^8\mathrm C_k}$ counts ways to select $k$ from $8$ different nuts.   So since box#1 can contain none to eight of them; we sum for all the ways to select which of how many.   $\tiny {^8\mathrm C_0}+{^8\mathrm C_1}+{^8\mathrm C_2}+{^8\mathrm C_2}+{^8\mathrm C_4}+{^8\mathrm C_5}+{^8\mathrm C_6}+{^8\mathrm C_7}+{^8\mathrm C_8}$

$2^8$ counts ways to make $8$ independent choices of $1$ from $2$ boxes.

They are just different views of the same task, so the counts must be equal.