Ben needs to form a committee of 3 from a group of 8 engineers to study design improvements for a product. If two of the
engineers are too inexperienced to serve together on the committee, how many different committees can Ben form?
20
30
50
56
336
Form a committee of 3 from 8 - 8C3 = 56
Exclude groups that have 2 engineers together - If say 1,2 need to be excluded - (1,2,3), (1,2,4), (1,2,5), (1,2,6), (1,2,7), (1,2,8) - 6 - PART 2
56-6 = 50
Is there any formula for such problems, where you can compute part 2 with ease?
Since its a small group, its fairly easy to compute, but if its big, it gets difficult.
Anybody?
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crimson2283
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maihuna
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Assuming the two incapable engg are tagged bb, the input is : xxxxxxbb
W/o any constraint : 8c3 = !8/!3*!5 = 8*7*6/3*2 = 56
Out of these combination where both b are there can be done in : 2b only one way, Out of 6x, one x in 6c1 = 6 ways
So reud no = 56-6 = 50
W/o any constraint : 8c3 = !8/!3*!5 = 8*7*6/3*2 = 56
Out of these combination where both b are there can be done in : 2b only one way, Out of 6x, one x in 6c1 = 6 ways
So reud no = 56-6 = 50
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manpsingh87
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when no inexperience engineer is selected total no. of ways would be 6C3 =20;crimson2283 wrote:Ben needs to form a committee of 3 from a group of 8 engineers to study design improvements for a product. If two of the
engineers are too inexperienced to serve together on the committee, how many different committees can Ben form?
20
30
50
56
336
Form a committee of 3 from 8 - 8C3 = 56
Exclude groups that have 2 engineers together - If say 1,2 need to be excluded - (1,2,3), (1,2,4), (1,2,5), (1,2,6), (1,2,7), (1,2,8) - 6 - PART 2
56-6 = 50
Is there any formula for such problems, where you can compute part 2 with ease?
Since its a small group, its fairly easy to compute, but if its big, it gets difficult.
Anybody?
when one of the inexperience engineer is selected total no. of ways would be 2C1 * 6C2 =30;
hence total no. of ways would be 30+20=50
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Jeff@TargetTestPrep
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We can use the following formula:crimson2283 wrote:Ben needs to form a committee of 3 from a group of 8 engineers to study design improvements for a product. If two of the engineers are too inexperienced to serve together on the committee, how many different committees can Ben form?
20
30
50
56
336
Total number of ways to select the committee = # of ways with both inexperienced engineers selected + # of ways with two inexperienced engineers NOT selected
Thus:
# of ways with two inexperienced engineers NOT selected = Total number of ways to select the committee - # of ways with both inexperienced engineers selected
Total number of ways to select the committee:
8C3 = (8 x 7 x 6)/3! = 56 ways
Now let's calculate the total number of ways to select the committee such that the two inexperienced engineers are both selected. One such occurrence would be:
NNE (N = inexperienced engineer and E = experienced engineer)
Since both inexperienced engineers have been selected, there is only 1 position left, and there are 6 experienced engineers to fill it. Thus, we have:
6C1 = 6
Thus, the number of ways to select the committee members with both inexperienced engineers NOT selected = 56 - 6 = 50.
Answer: C
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