A group of 6 people is to be chosen from 4 men and 11 women.
(a) In how many different ways can a group of 6 be chosen if it must contain exactly 1 man?
Two of the 11 women are sisters Jane and Kate.
(b) In how many different ways can a group of 6 be chosen if Jane and Kate cannot both be in the group?
Solution
(a) To choose a group of 6 with exactly 1 man, we select 1 man from 4 men and 5 women from 11 women. The number of ways to do this is given by:
\(\binom{4}{1} \times \binom{11}{5} = 4 \times 462 = 1848\)
(b) We need to find the number of ways to choose a group of 6 such that Jane and Kate are not both included. We can use two methods:
Method 1: Consider the scenarios:
- Neither Jane nor Kate is selected: Choose 6 from the remaining 13 people.
- Only Jane is selected: Choose 5 from the remaining 13 people.
- Only Kate is selected: Choose 5 from the remaining 13 people.
The total number of ways is:
\(\binom{13}{6} + \binom{13}{5} + \binom{13}{5} = 1716 + 1287 + 1287 = 4290\)
Method 2: Calculate the total number of ways to choose 6 people from 15, then subtract the number of ways where both Jane and Kate are included:
\(\binom{15}{6} - \binom{13}{4} = 5005 - 715 = 4290\)
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