Question

James and Peter are active members of the Student Guild and are graduating in a class of 27 students in total. A committee of five members to graduation must be formed to organize the festivities. How many committees can be formed so that James and Peter are members?

Answer 1

Well
Total number of students=27
Number of class students for a party=5
So number of committees that can be formed=27/5 = 5 at the maximum.

Answer 2

No tI got this!

Answer 3

Since both James and Peter have to be chose, that leaves 25 students left, and 3 spots on the committee left. Thus, the answer should be\[\binom{25}{3}\]or 25-choose-3

Answer 4

mehhh king got there first.. i was about to post that.. lol

Answer 5

.............

Answer 6

@How many parties can be formed if James and Peter ARE INCLUDED in each group or?

Answer 7

Wait, so each group requires 5 members,
James and Peter are members already so, 27-2=25 members
Moreover, each group would then have 5-2=3 members so,
it can be the way KingGeorge described it, though I'm not sure about if its right..anyways.

Answer 8

veng. stop..it..the king won..lol

Answer 9

He won but that's not stopping me from celebrating his victory, is it? :)

Answer 10

I always win..

Answer 11

whats the problem, hershey? veng helped me understand

Answer 12

There are 27 members in all, including James and Peter.
So, let first group have 3+James+Peter. I still think that only 5 groups can be formed with James and Peter being in one of them.

Answer 13

Hey King, by that you mean: 25!/3!22! ? cuz the result is huge

Answer 14

\[\binom{25}{3}= {25! \over {3!}(25-3)!}={25! \over {3! 22!}}\]And that is a very huge number.

Answer 15

Well not too huge actually, only 2300

Answer 16

@King: They haven't asked how many ways can the commities be formed so you sure that's a viable answer?

Answer 17

2300 sounds weird

Answer 18

2300 is correct because....(wait for next post)

Answer 19

out of those 25 people left to choose from, let's number them from 1 to 25. We can only choose three of them. That means, we have 25 choices for the first person, 24 for the next and 23 for the last. \[25*24*23 = {25! \over 22!}\]Then, each of these 3 person groups can be ordered 3! different ways, so you have to divide by 3!\[{25! \over 22!} \times {1 \over 3!} = {25! \over 3! 22!} = 2300\]

Answer 20

Right, that's the formula for finding the number of combinations which can be possible for the groups..
@Mel: How much marks is this question for?

Answer 21

Yep, I get it now. Thanks, King! mmm, not sure what you mean, ven.

Answer 22

Nevermind, guess it was the number of combinations possible for the groups..misunderstood the question lol.

Answer 23

Yea, me too. But that's correct, I have the options here :P
check them out:

a) 9288 b) 3276 c) 9828 d) 2300

Answer 24

Yeah well 2300 is already proved to be the answer :P

Answer 25

Yes, what I meant is that they are all ''huge'' numbers.