Funny party

Probability Level 3

Manon organizes a party and sends 13 invations to 4 women and 9 men. She doesn't know who will come but she is sure that:

There will be 10 people in total.
There will be at least 3 women.
There is a specific woman who will not come if a specific man comes (and she knows if he will).

How many ways the group for this party can be formed?

Details and Assumptions : Here, I'm asking for the number of groups which can be formed with the conditions stated. Manon is not part of the party, we consider the different groups without counting her .

The answer is 88.

1 solution

Chew-Seong Cheong
Aug 6, 2018

Let the specific women be Ms X, who will not come if a specific man Mr Y comes. Let the number of way to choose among the 4 women be $N_w$ , and to choose among the 9 men be $N_m$ . Then the number of ways to choose among the people for case $n$ is $N_n=N_wN_m$ .

Case 1: Mr Y will turn up, then Ms X won't turn up and the way to choose among the women is $N_w = \dbinom 33 = 1$ and $N_m = \dbinom 86 = 28$ $\implies N_1 = 1 \times 28 = 28$ .
Case 2: Mr Y won't turn up and 3 women turn up. Then $N_w = \dbinom 43 = 4$ , $N_m = \dbinom 87 = 8$ and $N_2 = 4\times 8 = 32$ .
Case 3: Mr Y won't turn up and 4 women turn up. Then $N_w = \dbinom 44 = 1$ , $N_m = \dbinom 86 = 28$ and $N_2 = 1\times 28 = 28$ .

Therefore, the total number of different groups possible is $N=N_1+N_2+N_3 = 28+32+28 = \boxed{88}$ .

Funny party

The answer is 88.

1 solution

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