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Splendidgirl666
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Amazing! Thanks a lot. It is all clear now.
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+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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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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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.

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Bunuel: Please untag probability
Splendidgirl666
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
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Splendidgirl666
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

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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Splendidgirl666
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

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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Splendidgirl666
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

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
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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.
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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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I'm going to Assume that the 3 Brand M Computers are Indistinguishable an the 2 Brand N Computers are Indistinguishable



(1st) In how many ways can we Select the 5 Offices out of 7 Total to each receive 1 Computer


"7 choose 5" = 7! / (5!2!) = 21 Ways


AND

(2nd) For Each different Combination of 5 Offices Chosen, How many Ways can we Arrange ---- M - M - M - N - N ?


5! / (2!3!) = 10 Unique Arrangements for each group of 5 Offices Selected




(21) * (10) = 210 ways

-B-
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