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# A company bought for its 7 offices 2 computers of brand N

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A company bought for its 7 offices 2 computers of brand N [#permalink]  17 Jan 2012, 14:38
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A company bought for its 7 offices 2 computers of brand N and 3 computers of brand M. In how many ways could computers be distributed among the offices if each office can have only 1 computer.

A. 196
B. 210
C. 256
D. 292
E. 312
[Reveal] Spoiler: OA
Magoosh GMAT Instructor
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Re: A company bought for its 7 offices 2 computers of brand N [#permalink]  17 Jan 2012, 16:17
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Hi, there. I'm happy to help with this.

This problem has to do with combinations. Here's the general idea: if you have a set of n elements, and you are going to choose r of them (r < n), then the number of combinations of size r one could choose from this total set of n is:

# of combinations = nCr = (n!)/[(r!)((n-r)!)]

where n! is the factorial symbol, which means the product of every integer from n down to 1. BTW, nCr is read "n choose r."

In this problem, let's consider first the three computers of brand M. How many ways can three computer be distributed to seven offices?

# of combinations = 7C3 = (7!)/[(3!)(4!)] = (7*6*5*4*3*2*1)/[3*2*1)(4*3*2*1)]

= (7*6*5)/(3*2*1) = (7*6*5)/(6) = 7*5 = 35

There are 35 different ways to distribute three computers to 7 offices. (The massive amount of cancelling that occurred there is very much typical of what happens in the nCr formula.)

One we have distributed those three M computers, we have to distribute 2 N computers to the remaining four offices. How many ways can two computer be distributed to four offices?

# of combinations = 4C2 = (4!)/[(2!)(2!)] = (4*3*2*1)/[(2*1)(2*1)] = (4*3)/(2*1) = 12/2 = 6

For each of the 35 configurations of distributing the M computers, we have 6 ways of distributing the N computers to the remaining offices. Thus, the total number of configurations is 35*6 = 210. Answer choice = B

Notice, we would get the same answer if we distributed the N computers first.

Number of ways to distribute 2 N computers to seven offices = 7C2 = (7!)/[(2!)(5!)] = (7*6)/(2*1) = 21

Number of ways to distribute the 3 M computers to the remaining five offices = 5C3 = (5!)/[(3!)(2!)] = (5*4)/(2*1) = 20/2 = 10

Product of ways = 21*10 = 210. Same answer.

http://magoosh.com/gmat/2012/gmat-permu ... binations/

Does this make sense? Please let me know if you any questions on what I've said.

Mike
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Mike McGarry
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Re: A company bought for its 7 offices 2 computers of brand N [#permalink]  17 Jan 2012, 16:29
Amazing! Thanks a lot. It is all clear now.
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Re: A company bought for its 7 offices 2 computers of brand N [#permalink]  17 Jan 2012, 19:08
+1 for B.

I solved it in a slightly different way. I would love feedback on the logic.
Again, this is a comb problem. So I though of it as follows.

From the seven offices, there will be 5 'winners' (of PCs) and 2'losers'. The number of ways they could be winners and losers is 7!/(5!2!) = 21

For the five winners, the 5 computers can be distributed in 5!/(2!3!) ways = 10.

If you have trouble seeing this, think of anagrams NNMMM (so 5! ways of ordering them but divide by 2! because 2 Ns are similar and 3! because 3 'M's are similar.)

The number of different combinations is then 21*10 = 210 -> B.

Cheers!
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Re: A company bought for its 7 offices 2 computers of brand N [#permalink]  18 Jan 2012, 03:15
Expert's post
Halawi wrote:
+1 for B.

I solved it in a slightly different way. I would love feedback on the logic.
Again, this is a comb problem. So I though of it as follows.

From the seven offices, there will be 5 'winners' (of PCs) and 2'losers'. The number of ways they could be winners and losers is 7!/(5!2!) = 21

For the five winners, the 5 computers can be distributed in 5!/(2!3!) ways = 10.

If you have trouble seeing this, think of anagrams NNMMM (so 5! ways of ordering them but divide by 2! because 2 Ns are similar and 3! because 3 'M's are similar.)

The number of different combinations is then 21*10 = 210 -> B.

Cheers!

Yes, that's correct:
Picking 5 winning offices from 7 - $$C^5_7=21$$;
Different assignment of NNMMM to these offices: $$\frac{5!}{2!3!}=10$$ (which basically is # of permutations of 5 letters NNMMM out of which 2 N's and 3 M' are identical).

Total: 21*10=210.

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Re: A company bought for its 7 offices 2 computers of brand N [#permalink]  21 Nov 2014, 16:41
Hi,
I managed to arrive at the correct number , 210, but in different way, so I would like to ask the experts if Im right or just happend to be coincidence to have it 210 as answer.

the first method: baisicaly I made three groups, M group, N group, and neither M nor N. so to arrange three groups there are 6 different ways or 3!, after that there are 7 officies and five comuters to destribute , 3 M and 2N, so I got 7 oficies times 5 PC's = 35, and when multiply 35 with 6 I got 210.

second method: 7 officies and again 3 M and 2N, so 7!/(3! x 2!) = 210.

Again Im not sure if I'm correct, so I'm kindly asking for experts to give opinion . Thanks
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Re: A company bought for its 7 offices 2 computers of brand N [#permalink]  22 Nov 2014, 05:33
Expert's post
kzivrev wrote:
Hi,
I managed to arrive at the correct number , 210, but in different way, so I would like to ask the experts if Im right or just happend to be coincidence to have it 210 as answer.

the first method: baisicaly I made three groups, M group, N group, and neither M nor N. so to arrange three groups there are 6 different ways or 3!, after that there are 7 officies and five comuters to destribute , 3 M and 2N, so I got 7 oficies times 5 PC's = 35, and when multiply 35 with 6 I got 210.

second method: 7 officies and again 3 M and 2N, so 7!/(3! x 2!) = 210.

Again Im not sure if I'm correct, so I'm kindly asking for experts to give opinion . Thanks

The first approach does not make sense to me, while the second one is correct.
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Re: A company bought for its 7 offices 2 computers of brand N   [#permalink] 22 Nov 2014, 05:33
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