How many different teams can be formed ...?

[Fatima Hasan]

Homework Statement

One engineering group consists of 5 men and 7 women .
If 2 women refuse to be on the same team together , how many different project teams can be formed consisting of 2 men and 3 women ?

Homework Equations

-

The Attempt at a Solution

5 C 2 * 5 C 3 + (2 C 1 * 5 C 2) = 120
Can somebody check my answer please ?

 

[Dick]

Homework Statement

One engineering group consists of 5 men and 7 women .
If 2 women refuse to be on the same team together , how many different project teams can be formed consisting of 2 men and 7 women ?

Homework Equations

-

The Attempt at a Solution

5 C 2 * 5 C 3 + (2 C 1 * 5 C 2) = 120
Can somebody check my answer please ?

If there are 7 women in the group and you need 7 women in the team then you have to pick the 2 women who won't serve together. Is there a typo?

 

[Fatima Hasan]

and you need 7 women in the team

3 women

 

[tnich]

One engineering group consists of 5 men and 7 women .
If 2 women refuse to be on the same team together , how many different project teams can be formed consisting of 2 men and 7 women ?

Is this the original problem statement? I find it ambiguous. Does
"2 women refuse to be on the same team together"

mean
a) "no two women will work on the same team together" or
b) "two particular women refuse to work on the same team together"

I think
"how many different project teams can be formed consisting of 2 men and 7 women"

means
"how many different project teams can be formed if each team must consist of 2 men and 7 women",

but in that case under either option a) or b) the number of teams that can be formed is zero.

So perhaps
"how many different project teams can be formed consisting of 2 men and 7 women"

means
"how many teams of at least two people" can be formed from 2 men and 7 women".

(One person does not constituted a team.) In that case the question has different answers under options a) and b).

Is there a type in the problem statement?
Should "how many different project teams can be formed consisting of 2 men and 7 women"

read instead
"how many different project teams can be formed consisting of 5 men and 7 women"?

I see you have responded to @Dick while I was typing this. Could you please give us a complete and correct problem statement?

 

[Fatima Hasan]

how many different project teams can be formed consisting of 2 men and 7 women

7 is a typo .
my question is "How many different project teams can be formed consisting of 2 men and 3 women "

 

[Dick]

7 is a typo .
my question is "How many different project teams can be formed consisting of 5 men and 2 women "

Looking at your answer I would say it's "How many different project teams can be formed consisting of 2 men and 3 women" as you said in your first correction. If so your answer has correct parts. It looks like the first term is how many ways to choose the team excluding the 2 women who don't get along. The next term looks like the ways to choose the womens team with those two. But the second term doesn't have any choice of men!? It would really be best if you explained your thinking rather than me guessing it.

 

[Fatima Hasan]

Looking at your answer I would say it's "How many different project teams can be formed consisting of 5 men and 3 women" as you said in your first correction. If so your answer has correct parts. It looks like the first term is how many ways to choose the team excluding the 2 women who don't get along. The next term looks like the ways to choose the womens team with those two. But the second term doesn't have any choice of men!? It would really be best if you explained your thinking rather than me guessing it.

The first term is how many ways to choose 2 men (5C2).
The next term is how many ways to choose 3 women , either choosing 3 women without the 2 women who don't get along (5C3) or (+) choosing one women of them and (*) two others (2C1 * 5C2) .

 

[Dick]

The first term is how many ways to choose 2 men (5C2).
The next term is how many ways to choose 3 women , either choosing 3 women without the 2 women who don't get along (5C3) or (+) choosing one women of them and (*) two others (2C1 * 5C2) .

Right. But then BOTH of the ways of choosing the women should be multiplied by the way to choose the men, not just the first.