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Permutation /combination Quest

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Permutation /combination Quest

by jawad » Sat Aug 23, 2008 9:37 am
Hi ,
Pls help findings logic to below question


Ben needs to form a commitee of 3 from a group of 8 engineers. 2 Engineers are inexperienced to be serve together in this committe. How many different commitees can ben form .

20
30
50
56
356

regards , Jawad
Jawad Shah
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Re: Permutation /combination Quest

by sudhir3127 » Sat Aug 23, 2008 10:12 am
jawad wrote:Hi ,
Pls help findings logic to below question


Ben needs to form a commitee of 3 from a group of 8 engineers. 2 Engineers are inexperienced to be serve together in this committe. How many different commitees can ben form .

20
30
50
56
356

regards , Jawad
I go with C 50

a committee of 3 has to be formed and not more than 1 inexperienced engineer

thus the combinations are
6C3 ( no inexperience engineer ) + 6C2*2C1 ( 2 experience engineers and 1 non experience)

6C3+ 6C2*2C1

20 + 15*2

= 50

Hope it helps..
do let me know if u have any doubts
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by parallel_chase » Sat Aug 23, 2008 10:12 am
Here's what I did

total number of combination = 8C3 = 56

number of combination both engineers are taken together = 1*6 =6

total number of combinations both engineers not taken together = 56-6=50


Hence the answer is 50.

Let me know if anybody thinks otherwise.
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by pranavc » Sat Aug 23, 2008 10:46 am
I am going with 50. Is that the official answer?
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by ankijain » Sat Aug 23, 2008 12:38 pm
Haha.. what dyu mean official answer? Is it an official question? :P

U know, somehow i worked it like this..

Total = 8 ppl.
But only one of the two inexperienced fellows can be selected (at most) which means, technically speaking, we have to select 3 ppl outta 7 (we take inexperienced ppl and club them as one).

Thus we select as 7C3. And since the 2 inexperienced ones can interchange among themselves, we multiply the result by 2.
Answer = 70.

Problem is, that option does not exist. Can someone tell me where I am going wrong?
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by pranavc » Sat Aug 23, 2008 12:44 pm
Well, the questions states that both inexperienced people cannot be together on the team. By calculating 7C3, you are ruling the possibility of either of the two experience people being on the team. Does that make sense?
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Permutation /combination Quest

by jawad » Sun Aug 24, 2008 1:27 am
Hi Sudhir ,

Correct Ans is 50.

But can u pls help me understand ur logic of adding the 2-combinations.
Though ur ans is correct but i would like to see working of ur idea.

regards. jawad
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by nervesofsteel » Sun Aug 24, 2008 1:57 am
C is the answer

Let us find all the committees where the 2 inexperienced are together

If both are selected we have to select 1 more from rest of 6
it can be done in 6C1 ways...
total number of ways in which a committee can be formed is 8C3

thus the number of ways in which a committee can be formed where
both inexperienced are not together is

8C3 - 6C1 = 50
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