there are 7 boys and 6 girls willing to serve on a committee. how many 7-member committees are possible if a committee is to contain:
a) 3 boys and 4 girls
b) at least one member of each sex?
thank in advance!!
a) What is C(7,3) x C(6,4) ?
b) with no restriction --- C(13,7)
all girls = C(6,4)
all boys = C(7,3)
exclude the all boys/girs from the total.
i did them but i got the wrong answers
for (a), the answer must be 525 and
for (b), the answer must be 1715
i got the rite answer for letter b though, by doing my own method
oh and thanks :]
To find the number of possible committees, we can use combinations.
A) For a committee with 3 boys and 4 girls, we need to choose 3 boys out of 7 and 4 girls out of 6. The number of ways to choose k items from a set of n distinct items is given by the combination formula:
C(n, k) = n! / (k! * (n-k)!)
So, for this case, we have:
C(7, 3) * C(6, 4) = (7! / (3! * (7-3)!)) * (6! / (4! * (6-4)!))
Simplifying:
(7 * 6 * 5) / (3 * 2 * 1) * (6 * 5) / (4 * 3* 2 * 1) = 35 * 15 = 525.
Therefore, there are 525 possible committees with 3 boys and 4 girls.
B) For a committee with at least one member of each sex, we can consider two cases:
1) One boy and six girls:
C(7, 1) * C(6, 6) = 7 * 1 = 7.
2) Two or more boys and five or fewer girls:
Total number of possible committees - Number of committees with one boy and six girls:
Total = C(13, 7) - C(7, 1) * C(6, 6) = (13! / (7! * (13-7)!)) - 7.
Simplifying:
(13 * 12 * 11 * 10 * 9 * 8) / (7 * 6 * 5 * 4 * 3 * 2 * 1) - 7 = 13 * 2 - 7 = 26 - 7 = 19.
Therefore, there are 7 + 19 = 26 possible committees with at least one member of each sex.
In summary:
a) There are 525 possible committees with 3 boys and 4 girls.
b) There are 26 possible committees with at least one member of each sex.