Seven managers and eight sales representatives volunteer to attend a trade show. Their company can afford to send five people. In how many ways can they be selected if there must be at least one manager and one sales representative chosen?
Thanks
2007-07-05 12:22:21 · 4 answers · asked by Icobes 2 in Science & Mathematics ➔ Mathematics
I did C(7,1) x C(8,1) x C(13,3) which is apparently wrong!
2007-07-05 12:23:20 · update #1
Suppose we disregard the restriction that there must be at least one sales rep and at least one manager; then there would be C(15,5) ways. Of these, C(8,5) have no manager and C(7,5) have no sales rep. So the number of acceptable combinations is C(15,5) - C(8,5) - C(7,5) = 2926.
2007-07-05 13:20:11 · answer #1 · answered by jw 2 · 0⤊ 0⤋
The probability that there will be 1 rep and 1 mgr is 1
The remaining 3 members can be selected from 6 managers and 7 reps so the number of combinations is 13C3
= 13!/10!*3!= 13*12*11/6=286
2007-07-05 20:04:32 · answer #2 · answered by ironduke8159 7 · 0⤊ 0⤋
Here's the answer: let's call m- #of members of management chosen, and s - # of sales representatives. They can select (m=1, s=4) + (m=2, s=3)+ (m=3, s=2)+ (m=4, s=1)=
C(7,1)*C(8,4)+C(7,2)*C(8,3)+C(7,3)*C(8,2)+C(7,4)*C(8,1)=2926
2007-07-05 19:58:37 · answer #3 · answered by magicbright2007 1 · 0⤊ 1⤋
⥠thus they are 15 men;
C of 15 by 5 = 15*14*13*12*11/5! = a1;
C of 7 by 5 = 7*6*5*4*3/5! =a2;
C of 8 by 5 = 8*7*6*5*4/5! =a3;
⦠ways = a1-a2-a3 =2926;
2007-07-05 20:09:59 · answer #4 · answered by Anonymous · 0⤊ 0⤋