For each problem, display the equation used to form the answer. Leave combinatio
ID: 3675938 • Letter: F
Question
For each problem, display the equation used to form the answer. Leave combinations, permutations, factorials, and exponents intact. Good example: C(6, 3) - C(5, 2) = 10 Bad examples: 10 no equation 20 - 10 = 10 did not leave combinations intact If the final equation is too complex, introduce variables and break it down into two or more equations. You are encouraged to show additional workto indicate how you derived your equation. A club with 15 women and 12 men need to form a committee that consists of a president, a vice president, a secretary, and a treasurer. How many committees are possible... if the committee must have two women and two men? if the committee must have at least two men?Explanation / Answer
Given there are 15 women and 12 men,
Need to for a committee of 4 people( president,vice president, secretary, treasurer)
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a) If the committe must have two women and two men
two women from 15 women can be selected in C(15,2) ways
Two men from 12 men can be selected in C(12,2) ways
Total no of ways = C(15,2) * C(12,2) = [15 * 14/ (1 * 2) ] * [12 * 11 /(1*2)] = 6930 ways
b) If the committe must have atleast two men?
So No of ways = committe having (2 men,2 women) + committe having (3 men,1 women) +committe having (4 men, 0 women)
committe having (2 men,2 women)
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The 2 men can be selected in C(12,2) ways
the other 2 members should be women, and selected in C(15,2) ways
No of ways = C(15,2) * C(12,2) = 6930
committe having (3 men,1 women)
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No of ways = C(15,1) * C(12,3) = 3300
committe having (4 men,0 women)
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No of ways = C(12,4) = 495
Total no of ways = 6930+3300+495 = 10725 ways
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