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Help - Probabilty question 3 on test 25

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Help - Probabilty question 3 on test 25 [#permalink] New post 06 Apr 2010, 10:08
Question - 4 professors and 6 students are being considered for membership on a supervisory committee which must consist of 3 people. If the committee has to include at least 1 professor, how many ways can this committee be formed?

My answer was wrong, but I am trying to understand what I did wrong while solving the question. Please help!

My solution -

Since atleast one professor must be in the committee, I first selected one professor which can be done in 4 ways as there are 4 professor. So I have 4 in the first position.

___4___ ________ _______

Now I have a total of 9 people (10 - 1 professor) and 2 spots in the committee. This can be done in 9C2 ways.

9!
------- = 36
7! *2!

So total ways is

4 * 36 = 144

Please let me know where I went wrong

Tahnks!
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Re: Help - Probabilty question 3 on test 25 [#permalink] New post 06 Apr 2010, 10:27
I get the answer 100 ways

we have 4 pr and 6 st.

so, we can form the committee in 3 ways:
2 pr and 1 st = 2C4 * 6= 36
1pr and 2 st = 4 * 2C6 = 60
3 pr 3C4 = 4

Then 36+60+4=100

Am I right?

Last edited by fruit on 09 Apr 2010, 22:50, edited 1 time in total.
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Re: Help - Probabilty question 3 on test 25 [#permalink] New post 06 Apr 2010, 10:42
total ways to form a committe with atleast 1 professor = total ways to form committee(both professors and students combined) - total ways to form committe without any professor(only students) = 10c3 - 6c3 = 120 - 20 = 100
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Re: Help - Probabilty question 3 on test 25 [#permalink] New post 06 Apr 2010, 12:04
At least an event = All possibilities - No such event occurring
At least a prof = All possibilities - No prof

all possibilities: 10C3 = 10*9*8/6 = 120
No prof involved: 6 students only forming the committee: 6C3 = 20

Therefore, At least a prof = 120 - 20 = 100

Hope this helps
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Re: Help - Probabilty question 3 on test 25   [#permalink] 06 Apr 2010, 12:04
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