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maths 08 question #4 doubt

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maths 08 question #4 doubt [#permalink] New post 28 Oct 2009, 09:06
If 4 women and 6 men work in the accounting department, in how many ways can a committee of 3 be formed if it has to include at least one woman?

The official solution to the above problem is

Consider an unconstrained version of the question: in how many ways can a committee of 3 be formed? The answer is 10c3. From this figure we have to subtract the number of committees that consist entirely of men: 6c3 The final answer is 10c3-6c3=120-20=100.

However my doubt is why cant we solve it using below

no of ways=4c1*6c2+4c2*6c1+4c3 (sum of 3 diff cases)
1st case--->select 1 women out of 4 and 2 men out of 6 to make a team of 3
2nd case--->select 2 w and 1 m to make a team of 3
3rd case--->select all 3 members from 4 available women
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Re: maths 08 question #4 doubt [#permalink] New post 28 Oct 2009, 13:27
ggmatt wrote:
If 4 women and 6 men work in the accounting department, in how many ways can a committee of 3 be formed if it has to include at least one woman?

The official solution to the above problem is

Consider an unconstrained version of the question: in how many ways can a committee of 3 be formed? The answer is 10c3. From this figure we have to subtract the number of committees that consist entirely of men: 6c3 The final answer is 10c3-6c3=120-20=100.

However my doubt is why cant we solve it using below

no of ways = 4c1*6c2 + 4c2*6c1 + 4c3 (sum of 3 diff cases)
1st case--->select 1 women out of 4 and 2 men out of 6 to make a team of 3
2nd case--->select 2 w and 1 m to make a team of 3
3rd case--->select all 3 members from 4 available women


You can do that way too as both are same.

1. 10c3 - 6c3 = 100
2. (4c1 x 6c2) + (4c2 x 6c1) + 4c3 = 4x15 + 6x6 + 4 = 60 + 36 + 4 = 100
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Re: maths 08 question #4 doubt   [#permalink] 28 Oct 2009, 13:27
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