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

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Bunuel
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What is the number of three elements sets of positive integers {a, b, [#permalink] New post  19 Feb 2021, 08:45
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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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Re: What is the number of three elements sets of positive integers {a, b, [#permalink] New post  19 Feb 2021, 21:12
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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

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Re: What is the number of three elements sets of positive integers {a, b, [#permalink] New post  19 Feb 2021, 22:34
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ravigupta2912 wrote:
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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.


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Re: What is the number of three elements sets of positive integers {a, b, [#permalink] New post  29 Aug 2021, 06:51
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CrackverbalGMAT wrote:

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



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)
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Re: What is the number of three elements sets of positive integers {a, b, [#permalink] New post  19 Feb 2021, 21:54
CrackVerbalGMAT wrote:
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

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

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Re: What is the number of three elements sets of positive integers {a, b, [#permalink] New post  19 Feb 2021, 23:45
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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Re: What is the number of three elements sets of positive integers {a, b, [#permalink] New post  20 Feb 2021, 00:22
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ravigupta2912 wrote:
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.

Posted from my mobile device



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


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Re: What is the number of three elements sets of positive integers {a, b, [#permalink] New post  20 Feb 2021, 09:06
CrackVerbalGMAT wrote:
ravigupta2912 wrote:
Posted from my mobile device


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


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)
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