Question

A friend of mine is giving a dinner party. His current wine supply includes 8 bottles of zinfandel, 10 of merlot, and 12 of cabernet, all from different wineries. He will serve five bottles of the wines during dinner in a particular order.

(a) How many distinct sequences are there of serving any five wines?

(b) If the first two wines have to be zinfandel and the last three must either be merlot or cabernet, how many ways can this be done?

(c) If the wine sequence is formed randomly, what is the probability that none of the wines served is a zinfandel?

Answer 1

Here there are 8 bottles of zinfandel, 10 of merlot and 12 of cabernet all from different winerie.

He has to serve five boottles of the wines during dinner in a particular order.

(a) Now here we have to find the number of disticnt sequences for serving any five wines.

so here there are 8+ 10 + 12 = 30 different type of wines. So,, for first bottle there are 30 choices and for second bottle there are 29 different choices and so on.

so number of different choices = 30 * 29 * 28 * 27 * 26 = 17,100,720

(b) here first two wines are zinfandel and the last three are from other two types that means first one can be from any of the 8 type of zinfadel and second one is from there remaining 7. Simirly, from last 3 it can be from any of the 10 + 12 = 22 different type of wines

so total number of different choices = 8 * 7 * 22 * 21 * 20 = 517,440

(c) If the wine sequence is served randomly, total number of sequences = ³⁰C₅

and none of them iszinfandel thatmeans 5 bottles are selected from 22 bottles.

so total number of choices in that case = ²²C₅

Probability that none of them is zinfandel = ²²C₅ / ³⁰C₅ = 0.1845