But the answer is supposed to be 120.Can you explain why?

[mutineer123]

combination problem: A football team consists of 3 players who play in a defence...

Homework Statement

http://www.xtremepapers.com/CIE/International%20A%20And%20AS%20Level/9709%20-%20Mathematics/9709_s05_qp_6.pdf

7 a ii (if exactly one forward player must be included, together with any two others)

Homework Equations

The Attempt at a Solution

I don't understand why my answer is wrong. So before answering to what the actual answer is, FIRST I want you to explain it to me why my answer is not right.

Heres what I did:
The possible combinations are

(D,M,F)
(D,D,F)
(M,M,F)

I got
3C1 X 3C1 X 5C1 +
3C1 X 2C1 X 5C1 +
3C1 x 2C1 x 5C1
=105

 

[zooxanthellae]

Note that I don't have a lot of experience with these types of problems so I may be wrong. However, you might find this helpful.

If I'm understanding this question correctly, it's stipulating that we must pick one forward and then we pick two people who aren't forwards. So we start with having 5 choices for forward. Once we've made that choice, we are picking from a pool of 6 people (all of the non-forwards). Once we've made that choice, we are picking from a pool of 5 people (the remaining non-forwards).

So the number of combinations is (5)(6)(5) = 150?

 

[mutineer123]

Note that I don't have a lot of experience with these types of problems so I may be wrong. However, you might find this helpful.

If I'm understanding this question correctly, it's stipulating that we must pick one forward and then we pick two people who aren't forwards. So we start with having 5 choices for forward. Once we've made that choice, we are picking from a pool of 6 people (all of the non-forwards). Once we've made that choice, we are picking from a pool of 5 people (the remaining non-forwards).

So the number of combinations is (5)(6)(5) = 150?

Your answer is wrong, but it ironically helped me understand why the right answer is right!(the part where you said Once we've made that choice, we are picking from a pool of 6 people). The answer is 75. Like you said we select a forward first (5C1) then we select 2 from the 6(3 def 3 mid) in any order, i,e (6C2). Their product gives the correct answer. But I still do not get why my answer is wrong.

 

[zooxanthellae]

Oh, I see my error (and yours!). The order that the other players are picked in is irrelevant, but our solutions work as if the order matters. Dividing by two accounts for the irrelevancy of order and produces the correct answer, 75. This works because by dividing by two we are effectively throwing out half of the possibilities, i.e. the situations in which the order of the second two players is the only thing that differs.

In my case you can just divide the whole thing by 2, but in your case you only need to divide the second 2 expressions by 2. Both then produce 75.

 

[azizlwl]

5.6C2=75

1st. you can choose 5 ways.
2nd. you can choose 6 ways
3nd. you can choose 5 ways( since 1 remove used for second place)

Total 150 but f,m,d = f,d,m (2 different permutation, actually 1 group)
Total permutation=150/2= 75