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combi ques

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by abhaypratapsingh » Fri Aug 01, 2008 6:29 am
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?



* 36
* 60
* 72
* 80
* 100
select 1 women 4C1 and then selecting 2 from rest of 9 is 9C2 , so total= 4C1 + 9C2

But the answer is 100 .
What Am I doing wrong ?
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by abhaypratapsingh » Fri Aug 01, 2008 9:06 am
can someone pls help to explain?
thnks,
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by JDesai01 » Fri Aug 01, 2008 10:13 am
This is a tough question that I definitely would not have been able to do in 2 minutes, but I think I got it. I am no expert, but here is an attempt to explain the question.

You need to do several separate combinations:

The possible groupings will be 3W, 2W/1M, 1W/2M (W=women, M=man)

For 3W: 3 combination 4 = 4

For 2W/1M: There are 2C4 possibilities for 2W = 6. Each of those unique pairs of 2W can be paired with one of the 6 men: 6*6=36

For 1W/2M: There are 4 options for 1 woman and 2C6 possibilities for 2 men. 2C6=15. 15*4 = 60

So, adding 4 + 36 + 60 = 100

A question like that would really throw me off on timing. Would love to know if there is an easier way to do this question.
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by abhaypratapsingh » Fri Aug 01, 2008 4:01 pm
easier way :
Total no ways = 10C3 =120
total no ways includig only men = 6C3 = 20

total no ways including at least 1 woman = 120 - 20 = 100.
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Re: combi ques

by Ian Stewart » Fri Aug 01, 2008 4:14 pm
abhaypratapsingh wrote: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?



* 36
* 60
* 72
* 80
* 100
select 1 women 4C1 and then selecting 2 from rest of 9 is 9C2 , so total= 4C1 + 9C2

But the answer is 100 .
What Am I doing wrong ?
We can choose 10C3 committiees. 6C3 will include only men (i.e. will include no women). The answer must be:

10C3 - 6C3 = Total of committees - those with only men = number of committees with at least one woman = 120 - 20 = 100

edit: abhaypratapsingh got there first! Nice solution.
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by blaque831 » Sat Aug 02, 2008 7:05 am
Can someone explain to me what exactly 10c3 and 6c3, or any similar expression for that matter is? I'm just now starting to understand permutation/combination questions and have no idea what XcX/XcX (10C4) means. Thanks in advance...
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by prashant77 » Mon Aug 04, 2008 3:38 am
at least one women can b chosen in following 3 ways
1w+2m; 2w+1m; 3w
this can be done as
4c1*6c2+4c2*6c1+4c3
60+36+4
=100
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by gabriel » Mon Aug 04, 2008 5:38 am
moved to the PS section
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by Ian Stewart » Mon Aug 04, 2008 6:23 am
blaque831 wrote:Can someone explain to me what exactly 10c3 and 6c3, or any similar expression for that matter is? I'm just now starting to understand permutation/combination questions and have no idea what XcX/XcX (10C4) means. Thanks in advance...
"10C3" means "the number of ways to choose 3 things from a set of 10 if the order of the 3 things does not matter".

In general, nCr = n!/[(r!)(n-r)!], so 10C3 = 10!/[(3!)(7!)] = (10*9*8)/(3*2*1) = 120.

When you choose, say, a committee or a board of directors, the order does *not* matter- if you list the names you've selected in a different order, you still have the same committee or board. If you are choosing a seating arrangement, lining people up, or choosing people to fill positions- say President, Vice-President and Treasurer- then order *does* matter; choosing A for President and B for Vice-President is different from choosing B for President and A for Vice-President.
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