Counting Question?

Question

7 women and 9 men are on the faculty in the mathematics department at a a school.

a) How many ways are there to select a committee of five members of the department if at least 1 woman must be on the committee?

b) How many ways are there to select a committee of five members of the department if atleast 1 woman and at least 1 man must be on the committee?

**This is a sperate question.**
How many strings of six lowercase letters from the english alphabet contain.
a) the letter a? I got an answer of 26^6 - 25^6
b) the letters a and b? I got: 26^6 - 24^6
c) the letters a & b in consecutive positions with a preceding b, with all the letters distinct? I couldn't get this one.
d) the letters a & b, where a is somewhere to the left of b in the string, with all the letters distinct? I couldn't get this one either.

Thank you very much for any help!!!

Answer 1

a)
the number of ways of picking 5 from 16 is what the question is actualy asking minus combinations of 5 from 9 , because there must be 1 woman
so 5 from 16 is given by :
(16x15x14x13x12) / (5x4x3x2x1)
which is :
4 x 3 x 14 x 13 x 2 = 4368
combinations of 5 from 9 is :
(9x8x7x6x5) / (5x4x3x2x1)
which is :
3 x 2 x 7 x 3 x 1 = 126
so the final answer is 4368-126 = 4242

b)
like before work out all the combinations that are possible. Then take off the combinations with only women or only men
this is given by :
5 from 16 is 4368 (worked out earlier)
and 5 from 9 is 126 (worked out earlier)
and 5 from 7 is :
(7x6x5x4x3) / (5x4x3x2x1)
which is :
7x6 / 2 = 21
so the final answer is 4368-126 (only men) - 21 (only women)
which is : 4221

i may come back and answer the 2nd question latter

by the way the 2nd question dosen't specify if the strings of letters can include repeats or not, for example with repeats allowed abcdea would be allowed, but if repeats are not allowed then there would be less combinations.

Answer 2

Second question:

a and b are correct.

For c, there are 5 possible positions of the needed a,b pair.

Wherever those two letters go, there are 24x23x22x21 possibilities for the rest. So the answer is 5x24x23x22x21.

For d, it's the same idea. Only now the possible number of a,b pairs is 5+4+3+2+1 (depending on whether a is leftmost, second-leftmost, etc.).

As for the first question,

For a, there are 16 choose 5 overall possibilities. But 9 choose 5 are bad (because they have all men), so subtract that number off.

For b, it's the same idea, only now you have to subtract off 9 choose 5 for the men and also 7 choose 5 for the all-women possibility.