Question

Help me out
From 5 women nd 7 men, a committee of 2 women and 3 men is to be formed
(a) same number of persons but two men refused to work together
(b) rashed nd his girlfrnd always wants to work together

Answer 1

HI!!

Answer 2

Hi

Answer 3

\[\binom{5}{2}=\frac{5\times 4}{2}=10\]choices for the two women

Answer 4

Can u explain @misty1212

Answer 5

i am still thinking about the second part, 3 out of 7 men but two refuse to serve together

Answer 6

See i have solve it but donnow if its correct or not

Answer 7

if there were no restrictions on the men it would be \[\binom{5}{2}\times \binom{7}{3}\]
but we need to figure out how many ways the two men could be on the same committee so we can subtract that off

Answer 8

5C2[(5C2×2C1)+(5C3×2C0)]= 300

Answer 9

@misty1212
The two men who refuse to work together can be considered a single person, one or the other, or none at all.

Answer 10

so \(\binom{6}{3}\)?

Answer 11

yes!

Answer 12

@mathmate check my answer and methid

Answer 13

@misty1212 check my method

Answer 14

@mathmate check my answer and methid

Answer 15

5C2[(5C2×2C1)+(5C3×2C0)]= 300

Answer 16

See i have solve it but donnow if its correct or not

Answer 17

Hi

Answer 18

Can u explain @misty1212

Answer 19

@misty1212
oops, on second thought, (6,3) won't work, because we would be missing something.
It would be more like (5,3)+2(5,2) instead of (7,3).

Answer 20

@statisz
Are there three parts to the question?
the question itself, then (a), and (b)?

Answer 21

I answered part a

Answer 22

@statisz
The question is not very clear to me.
Are there three parts to the question?
the question itself, then (a), and (b)?
and
Is the condition in (a) (where the two men refuse to work together) included in (b)?

Answer 23

No a is not included in b

Answer 24

@statisz
The question without constraints (a) and (b) has been answered by @misty1212, i.e. (5,2)*(7,3), namely the product of number of ways of choosing the women (5 choose 2) and number of ways of choosing the men (7 choose 3).
ok so far?

Answer 25

Yup okay

Answer 26

And dn if two mens refuse to be togther dn wat to do

Answer 27

@mathmate help.me plz
Or tag someone who knows this

Answer 28

For part (b) there are two cases, either
1. Rached and his gf are in: then there are 1 woman and 2 other men left in the choices, namely (4,1)*(6,2), or
2. Rached and his gf are out: then there are 4 women and 6 men to choose from, making (4,2)*(6,3)

Add the two cases together to give part (b).

Part (a) will follow.

Answer 29

For case (a):
Start with (5,3), i.e. not choosing any of the trouble makers, then \(add\)
(2,1)*(5,2) by choosing one of the problematic men plus 2 other men.
The total is then (5,3) + 2(5,2)=30.
Another way to work is to include one of the problematic men, then there are (6,3)=20 ways. Including the 7th man would add C(1,1)*(5,2)=10 for a total of 30 again.

Answer 30

Tnx @mathmate