Leaving C(11,2)=55
c) This part is stumping me.
Would it be 7 men - 5 women=2 remaining?
I don't know how to continue.
Twelve graduate students have applied for three available teaching assistantships. In how many ways can the assistantships be awarded if:
a) No preference is given to any student?
b) One particular student must be awarded an assistantship?
c) The group of applicants includes seven men and five women and it is stipulated that at least one woman must be awarded an assistantship?
a) No preference, so I think that the combination formula would take place. C(12,3) = 220
b) If one must be awarded would I subtract 1 from the 12 grad students and the 3 assistantships?
Leaving C(11,2)=
2 answers
a) and b) are correct
for c) you could take the cases of 1 woman, 2 women and 3 women, but a simpler way is to come in the back door, that is, you don't want the case of all 3 being men.
From a) we know that with no restrictions there are C(12,3) or 220
Now take the case where all the 3 choices are men from the 7 men which is
C(7,3) = 35
So the number of cases with at least one women = 220 - 35 or 185
long way
1 woman -- C(7,2) x C(5,1) = 105
2 women -- C(7,1) x C(5,2) = 70
3 women -- C(7,0) x C(5,3) = 10 , for a total of 185
for c) you could take the cases of 1 woman, 2 women and 3 women, but a simpler way is to come in the back door, that is, you don't want the case of all 3 being men.
From a) we know that with no restrictions there are C(12,3) or 220
Now take the case where all the 3 choices are men from the 7 men which is
C(7,3) = 35
So the number of cases with at least one women = 220 - 35 or 185
long way
1 woman -- C(7,2) x C(5,1) = 105
2 women -- C(7,1) x C(5,2) = 70
3 women -- C(7,0) x C(5,3) = 10 , for a total of 185
1 answer