Number of ways John and Mary can each select $4$ ice-cream flavours at two stores that sell one common flavour

Question

John and Mary are going to host an ice-cream party. They want to provide $8$ different flavours. John goes to store A to buy the first $4$ ice cream and Mary goes to store B to buy the other $4$ ice-cream. Assume that there are $7$ different flavours sold in each store; and store A and store B share $1$ similar flavour (ex:store A and Store B both have the Vanilla flavour). Calculate the number of possible ice- cream combinations when they buy the ice cream! (Different orders are not considered as different combinations)

The problem is I can't find the repetitive part. I think it is just $C(7,4) \cdot C(7,4)=1225$. Even I count it in different cases, it still adds up to be $1225$.

Answer 1

The invalid cases, instead of repetition, of $\binom74\binom74$ are when both people buy the common flavour (i.e. vanilla).

The number of cases that this happens is $\binom63\binom 63$. So the answer is

$$\binom74\binom74-\binom63\binom63 = 1225-20^2 = 825$$

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The additive and subtractive answers here show the identity:

$$\begin{align*} \binom nr^2 &= \left[\binom{n-1}{r-1}+\binom{n-1}r\right]^2\\ &= \binom{n-1}{r-1}^2 + 2\binom{n-1}{r-1}\binom{n-1}r + \binom{n-1}r^2 \end{align*}$$

In particular, $n=7$ and $r = 4$.

Answer 2

The problem with $\binom{7}{4} \cdot \binom{7}{4}$ is that you include all cases where they both end up buying the same ice cream. We want to avoid those cases, or John and Mary just need to communicate a bit better ;)

The comment provided by Peterwhy gave a good case distinction:

Case 1: John definitely buys the common flavour, so Mary shouldn't: $ 1 \cdot\binom{6}{3} \cdot \binom{6}{4}$

Here the $1$ represents vanilla, then there are 6 flavours left from which John can pick $3$ each. Mary has to still pick $4$ flavours, but not vanilla.

Case 2: John does not buy the common flavour, so Mary should: $1 \cdot \binom{6}{3} \cdot \binom{6}{4}$ (symmetric)

We don't really care about John or Mary, they're different people, but as long as we get our ice cream!

Case 3: Neither picks the common flavour: $\binom{6}{4} \cdot \binom{6}{4}$

oh well, we can live without vanilla I guess. If we add this all up we get $2 \cdot \binom{6}{3} \cdot \binom{6}{4} + \binom64 ^2= \binom64(2 \binom63 + \binom64)=15 \cdot (2\cdot 20+15)=15 \cdot(55)=825.$