Q. A student committee will be formed from different leaders of student groups. Exactly 1 of the 11 Fraternity presidents will be on the committee and exactly 2 of the 8 sorority presidents will be on the committee. Also, exactly 3 of the 7 leaders of the student activist groups will be included on the committee and no student holds more than one position and all student leaders are eligible. How many different committees may be formed?
(A) 616
(B) 1400
(C) 10,780
(D) 47,040
(E) 230,230
Given that three separate populations are used to form the committee, the answer will be the product of the numbers of possibilities of each group. What numbers will the answer have to be multiples of? Which answers aren't a multiples of those numbers? Based on the size of the different populations is there another factor that the answer must be a multiple of?
Solution
[spoiler]Since there is only one fraternity president on the committee, the answer must be a multiple of 11. Only answers (A), (C) and (E) are.
When picking a group of 3 from a population of 7, the combination will have to be a multiple of 5, since nothing in the denominator of the calculation will be a multiple of 5. This means (A) cannot be right since it is not a multiple of 5.
Finally, with a group of 2 from a population of 8, the combination will be a multiple of 4, because the numerator will have seven powers of 2 (8,6,4,2) but the denominator will only have five (6,4,2,2). This means (E) cannot be right since it is not a multiple of 4.
(C) is the correct answer.[/spoiler]
(A) 616
(B) 1400
(C) 10,780
(D) 47,040
(E) 230,230
Given that three separate populations are used to form the committee, the answer will be the product of the numbers of possibilities of each group. What numbers will the answer have to be multiples of? Which answers aren't a multiples of those numbers? Based on the size of the different populations is there another factor that the answer must be a multiple of?
Solution
[spoiler]Since there is only one fraternity president on the committee, the answer must be a multiple of 11. Only answers (A), (C) and (E) are.
When picking a group of 3 from a population of 7, the combination will have to be a multiple of 5, since nothing in the denominator of the calculation will be a multiple of 5. This means (A) cannot be right since it is not a multiple of 5.
Finally, with a group of 2 from a population of 8, the combination will be a multiple of 4, because the numerator will have seven powers of 2 (8,6,4,2) but the denominator will only have five (6,4,2,2). This means (E) cannot be right since it is not a multiple of 4.
(C) is the correct answer.[/spoiler]
