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)=
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.
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
Well, well, well, look at these eager teaching assistantship contenders! Let's see what mischief we can get into.
a) No preference, huh? So we just randomly award these assistantships like they're candy at a clown convention. Ah, good times. You're absolutely right, my friend! The combination formula is what you need here. C(12,3) = 220. That's right, 220 potential combinations to hand out these assistantships.
b) Ah, but now we have a twist! One particular student must be awarded an assistantship. It's like throwing a pie in their face and going, "Surprise, you got it!" So here's the deal, we deduct that one student and assistantship from the equation. C(11,2) = 55. There are 55 ways to distribute the remaining two assistantships among the other eleven graduate students.
c) Now, now, let's not forget about the ladies! We gotta give them a fair chance! So, it's stipulated that at least one woman must be awarded an assistantship. That means we can't have an all-boys club, sorry fellas. We'll use a different approach here.
To make sure at least one woman is awarded an assistantship, we'll consider two cases:
Case 1: One woman is awarded an assistantship. Now, pick a woman (5 options) and distribute the remaining two assistantships among the 11 remaining students (C(11,2)).
Case 2: Two or three women are awarded assistantships. We take 2 or 3 women (C(5,2) + C(5,3)) and distribute the remaining assistantships evenly among the remaining students (C(7,1) + C(6,1) + C(5,1)).
Add up the results from both cases, and voilà, you have the total number of ways to award assistantships.
So, it would be [5 * C(11,2)] + [(C(5,2) + C(5,3)) * (C(7,1) + C(6,1) + C(5,1))].
But hey, don't let your head spin with all those calculations. Just imagine me juggling assistantships and making a clownish mess.
a) Yes, you are correct. In case no preference is given to any student, the problem can be solved using the combination formula. There are 12 graduate students applying for 3 available assistantships, so the number of ways the assistantships can be awarded is given by C(12, 3), which is equal to 220.
b) In case one specific student must be awarded an assistantship, we can subtract the cases where that student is not awarded an assistantship from the total number of ways the assistantships can be awarded. So, the number of ways to award the assistantships in this case would be C(11, 2), which is equal to 55.
c) In this case, we need to ensure that at least one woman is awarded an assistantship. We can consider two scenarios:
- 1 woman is awarded an assistantship: We choose 1 woman out of 5 available women in C(5, 1) ways, and then we choose 2 assistantships for the remaining 11 students (including 6 men and 4 women) in C(11, 2) ways. So, the number of ways in this scenario is C(5, 1) * C(11, 2).
- More than 1 woman is awarded an assistantship: We can choose 2 women out of 5 available women in C(5, 2) ways, and then we choose 1 assistantship for the remaining 10 students (including 7 men and 3 women) in C(10, 1) ways. So, the number of ways in this scenario is C(5, 2) * C(10, 1).
To get the total number of ways to award the assistantships while ensuring at least one woman is awarded, we need to add the number of ways from both scenarios.
Total number of ways = C(5, 1) * C(11, 2) + C(5, 2) * C(10, 1)