In forming a committee of 5 out of 5 males and 6 females how many choi...
5 males and 6 females = 11 people. A committee of 5 out of 11 people can be selected in 11C5 =462 ways. At least female basically means that it can’t be an all-committee group. In this case, there is only 1 way to make an all-male committee, so just remove that case and you have 462–1=461 ways to choose a committee with at least one female.
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In forming a committee of 5 out of 5 males and 6 females how many choi...
To form a committee of 5 out of 5 males and 6 females, we need to find the number of choices we have to make so that there are 3 males and 2 females.
To select 3 males out of 5, we can use combination formula:
C(5,3) = 5! / (3! * 2!) = 10
To select 2 females out of 6, we can use combination formula:
C(6,2) = 6! / (2! * 4!) = 15
To form a committee of 5 with 3 males and 2 females, we need to multiply these two numbers:
10 * 15 = 150
Therefore, we have 150 choices to make to form a committee of 5 with 3 males and 2 females.
If we need to form a committee with at least one female, we can use the complement rule:
Total number of ways to form a committee of 5 out of 11 people = C(11,5) = 11! / (5! * 6!) = 462
Number of ways to form a committee with no females = C(5,5) = 1
Therefore, number of ways to form a committee with at least one female = 462 - 1 = 461
Hence, the correct answer is option D.