What is the number of three elements sets of positive integers {a, b,

Question

What is the number of three elements sets of positive integers {a, b, c} such that a × b × c = 2310 ?

A. 37
B. 38
C. 39
D. 40
E. 41

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CrackverbalGMAT · Accepted Answer

2310 broken into its factors = 1 * 2 * 3 * 5 * 7 * 11

Let us take the 5 factors 2,3,5,7 and 11 and find the number of sets possible

Taking 2 - 2 - 1 = 5C2 * 3C2 * 1C1 / 2! = 15 ways

Taking 1 - 1 - 3 = 5C1 * 4C1 * 3C3 / 2! = 10 ways

Taking 2 - 3 and the last number being 1. 5C2 * 3C3 = 10 ways

Taking 4 - 1 and the last number being 1. 5C4 * 1C1 = 5 ways

Total number of ways = 15 + 10 + 10 + 5 = 40 ways

Option D

Arun Kumar

ravigupta2912 · Answer

Hi, could you explain why you divided by 2!?

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CrackverbalGMAT · Answer

If we have say 4 letter a, b, c, d where we have to choose 2 - 1 - 1 letters

So we would have 4C2 * 2C1 * 1C1 = 12 ways

These 12 ways would be

ab, c, d and ba, c , d

ac, b, d and ca, b , d

ad, b, c and da, b , c

bc, a, d and cb, a , d

bd, a, c and db, a , c

cd, a, b and dc, a , b

The reverse is the same, so we divide by 2! = 2 ways

If we have something like 7C2 * 5C2 * 3C1 * 2C1, then the number of ways would be divided by 2!2!. We basically look at the subscript to see how many are the same and divide by that number of factorial to eliminate the repeated combinations.

Arun Kumar

ravigupta2912 · Answer

Okay. I think I was getting confused since combinations essentially mean that order doesn’t matter.

Shouldn’t the formula for combinations naturally take care of repeated arrangements?

I’m a little confused.

If it was a normal question of choosing 2 out of 4, we wouldn’t have done the division.

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CrackverbalGMAT · Answer

If it was simply 4C2, then you will not have to consider repetitions. You would get 6 ways = ab, ac, ad, bc, bd and cd

Suppose you have 4 letters and you want to choose 3 and then 1, you will get 4C3 * 1C1 = 4 ways

These are abc and d, abd and c, acd and b, bcd and a. Here again we do not have to worry about repetitions.

But when choosing from the same set, if the same amount is chosen we need to divide by the factorial. For eg if 7 letters are there and you want to choose 7C2 * 5C2 * 3C2 * 1C1, then this will have to be divided by 3!

Hope this helps

Arun Kumar

reii1998 · Answer

Hi, after reading this comment I understood why we should divide, but then in the main problem you wrote:
Taking 2 - 2 - 1 = 5C2 * 3C2 * 1C1 / 2! = 15 ways

Shouldn't we divide by 2!2! since we multiply 5C2 by 3C2 ? (2-2-1)

kaladin123 · Answer

CrackverbalGMAT  - Thank you for posting the solution. Can you please help me understand why {2310,1,1} is not a valid case? (i.e. Taking 5 numbers at once, the last two numbers being 1)

MuszerN · Answer

The question asks for three elements sets of positive integers. Per definition a set contains only unique elements; no element is allowed to be repeated. Therefore, {2310,1,1} is not a valid set.

Danush649 · Answer

@Bunuel@chetan2u
I have a doubt 
the factors are 1,2,3,5,7,11
So we have to arrange these 6 factors in _ *_ * _ right?
If a,b,c are unique 
why would 8!/(6!*2!) not work and if a,b,c dont need to be unique then there is just one more way 1 * 1* 2310

Im not able to tell what im missing here

deletus · Answer

The question is saying three element sets, yet you have a set of 2-3 and 4-1. Are these not two element sets? I am not understanding how you still got to the right answer.

once we get to the 5 factors, couldn't we just do 5C2*3C2 + 5C3 = 40 
5C2*3C2 for the 2-2-1 and 5C3 for the 3-1-1. This should work because we are not double counting anything. Am I mistaken?

ZIX · Answer

"... yet you have a set of 2-3 and 4-1. Are these not two element sets?"

Their 3rd element is 1, which was omitted at step 1 by considering only 5 factors (instead of 6)

ZIX · Answer

"Shouldn't we divide by 2!2! since we multiply 5C2 by 3C2 ? (2-2-1)"

No. We check how many same groups are being chosen, and then we divide by this number!

So, in 5  C2   * 3  C2   * 1C1 you are choosing 2 groups with same number of slots, so you divide by 2! for overcounting.

Second example, 7  C2   * 5  C2   * 3  C2   * 1C1: here we are choosing 3 groups with same number of slots, so you divide by 3! for overcounting.

third example, 7  C2   * 5  C2   * 3  C1   * 2  C1  : here we are choosing 2-2 groups with same number of slots, so you divide by 2!2!.

Akarsh97 · Answer

This will explain the need to diviide the combinayion of (3,1,1) and (2,2,1) by 2.

What is the number of three elements sets of positive integers {a, b,

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What is the number of three elements sets of positive integers {a, b,

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