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.

Here's a blog article I wrote about this topic, with some practice questions.

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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Amazing! Thanks a lot. It is all clear now.

+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.

Answer: B.
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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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If I solve the problem in following way, why it would be wrong?

1st computer can be distributed in 7 offices, i.e in 7 ways
2nd computer can be distributed in remaining 6 offices, i.e in 6 ways
3rd computer can be distributed in remaining 5 offices, i.e in 5 ways
4th computer can be distributed in remaining 4 offices, i.e in 4 ways
5th computer can be distributed in remaining 3 offices, i.e in 3 ways

So the total # of distribution is 7*6*5*4*3 = 2520.

I know, the above solution is not correct. I have posted it to identify my conceptual deficiency.

Moreover, under the same brand, are all computers same? Since, all computers have their own unique id or product code.

Kudos have been kept reserved for the right respondents.
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Bunuel: Please untag probability
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Select 5 offices for 5 computers = 7C5
Select 2 offices out of 5 selected offices for N type computer = 5C2
Remaining 3 computers will automatically be assigned to remaining 3 offices

Total Ways to assign computers = 7C5 * 5C2 = 210

ANswer: Option B
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Since there are 2 + 3 = 5 computers, 5 offices will have computers and 2 offices will not have computers. Let’s first determine the number of ways the offices that will have computers can be selected. We can choose 5 offices from an available 7 offices in 7C5 = (7 x 6)/2 = 21 ways.

Next, given a choice of 5 offices, let’s determine in how many ways we can distribute the 2 brand N computers and 3 brand M computers. Using the permutation formula with indistinguishable items, we see that this is possible in 5!/(3! x 2!) = (5 x 4)/2 = 10 ways.

Thus, the total number of distributions is 21 x 10 = 210.

Answer: B
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total computers 5 which can be distributed among 7 offices in 7c5 ; 21 ways
and total ways to distribute 2 & 3 computers ; 5!/2!*3! ; 10 ways
21*10 ; 210
OPTION B

I solved it this way. Bunuel, please review and let me know if this is correct.

There are 7 offices. 3 of these will have M computers, 2 will have N and 2 will have none, say O.

So, the total ways we can arrange MMMNNOO = 7! / 3!2!2! = 210.

Total Offices: 7

Total computers: 5

The number of ways to distribute 5 computers among 7 offices: \(^7{C_5}\) = 21 ways.

The number of ways 2(N) and 3(M) brands can be distributed in 5 offices: \(\frac{5! }{ (3! * 2!)}\) = 10 ways.

Total ways: 21 * 10 = 210

Answer B
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