Self Doubt

Question

From a group of 5 woman and 7 man we have to select a committee consisting of 2 woman and 3 men. Find the total number of ways to select such committed if (1 and 2 are a separate question)
1. Four man refuse to be in the same committee
2. 2 woman refuse to be in the same committee.

Comments

1.

i)Out of the 4 men we can select only one man in 4C1 way and from remaining 3 men we can select 2 men in 3C2 ways,

So total ways of forming the committee is 4*3*5C2=12*10=120

ii)Out of the 4 men if no one is in the committee then from remaining 3 men we can select 3 men in 3C3 ways.

So total ways to forming the committtee is 1*5C2=10

So answer is 120+10=130

2.

i)Selecting only one woman out of 2 is 2C1

Total number of ways=2C1*3C1*7C3=6*35=210

ii)Selecting no woman out of the 2 is 2C0=1

Total number of ways=3C2*7C3=105

So answer is 210+105=315.

Please correct me if i am wrong.

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it is correct answer

Answer 1

1.

i)Out of the 4 men we can select only one man in 4C1 way and from remaining 3 men we can select 2 men in 3C2 ways,

So total ways of forming the committee is 4*3*5C2=12*10=120

ii)Out of the 4 men if no one is in the committee then from remaining 3 men we can select 3 men in 3C3 ways.

So total ways to forming the committtee is 1*5C2=10

So answer is 120+10=130

2.

i)Selecting only one woman out of 2 is 2C1

Total number of ways=2C1*3C1*7C3=6*35=210

ii)Selecting no woman out of the 2 is 2C0=1

Total number of ways=3C2*7C3=105

So answer is 210+105=315.

Comments

why did we multiply it with 7C3 in the ii part?