For how many three-element sets of positive integers {a, b, c} is it t

Question

For how many three-element sets of positive integers {a, b, c} is it true that a × b × c = 2310?

(A) 32

(B) 36

(C) 40

(D) 43

(E) 45

Funkenflug · Answer

This is a very nice question, but I believe there is an error. If I assume that there are 5 prime numbers to recombine, I get C. However, there should be 6 prime numbers - 1,2,3,5,7,11. Could it be that the 1 was forgotten when writing the question?

Posted from my mobile device

Posted from my mobile device

aayush.works · Answer

Could you elaborate as to how you arrived at C ?

jnitish25 · Answer

2310= 2*3*5*7*11. Considering the set {a,b,c} such that a*b*c = 2310 , the 5 prime numbers can be divided into a group of 3 different numbers in either of the ways : 
(multiple of 2 numbers)*(multiple of 2 numbers)*(multiple of 1 number) = 5c2*3c2*1=30
or
(multiple of 3 numbers)*(multiple of 1 number)*(multiple of 1 number) = 5c3*1*1=10
So, 30+10 =40 sets.

srini2117 · Answer

1 is not a prime number

Funkenflug · Answer

You're absolutely right, I misspoke when I referred to it as a prime number. It is, however, still necessary to include it as a factor - for example, 1*1*2310 is a correct answer and including it yields more than 40 sets.

nick1816 · Answer

You are right!! It's not mentioned in the question that a,b or c can't be equal to 1.

briantoth6 · Answer

I'm getting something else entirely, I get 65.

Logic (similar to the methodology used to solve assuming a,b,c were all prime):

I found 4 types of a,b,c combos to divide the 6 #'s (1,2,3,5,7,11), being careful with the 1 value, because I think it can create complications with the methodology at the very end of my response.

1.) 1, product of three #'s, product of two #'s (e.g. {1,30,77})

1*5C3*2C2 = 1*10*1 = 10

2.) 1, product of four #'s, one remaining # (e.g. {1,210,11})

1*5C4*1C1 = 1*5*1 = 5

3.) product of three #'s, one remaining #, one remaining # (e.g. {30, 7, 11})

5C3*2C1*1C1 = 10*2*1 = 20

4.) product of two #'s, product of two #'s, one remaining # (e.g. {30, 7, 11})

5C2*3C2*1C1 = 10*3*1 = 30

10 + 5 + 20 + 30 = 65

~~~~~~~~~~~~~~~~~~~~~~~~~~

I tried doing it as though it was just 6 #'s (see below work) and the 1 value, I believe, messes things up.

product of 4#'s * 1# * 1# = 6C4*2C1*1C1 = 15*2*1 = 30

product of 3#'s * product of 2#'s * 1# = 6C3*3C2*1C1 = 20*3*1 = 60

product of 2#'s * product of 2#'s * product of 2#'s = 6C2 * 4C2 * 2C2 = 15*6*1 = 90

30 + 60 + 90 = 180

KarishmaB · Answer

2310 = 2*3*5*7*11

So we have 5 distinct numbers which need to be combined to give 3 groups.

Each number needs to be allocated to the 3 elements (a, b and c) such that no two elements are the same. The elements cannot be the same except 1, 1 so we need to avoid that

(4, 1, 0) - Put 4 factors together and 1 separately. The factor that is put separately can be picked in 5 ways. e.g. (1, 2, 1155)

(3, 1, 1) - Put 3 factors together and 2 separately. Pick 3 of the 5 factors to be put together in 5C3 = 10 ways. e.g. (7, 11, 30)

(3, 2, 0) - Put 3 factors together and 2 together. Pick 3 of the 5 factors to be put together in 5C3 = 10 ways. e.g. (30, 77, 1)

(2, 2, 1) - Put 2 factors together, another 2 together and 1 separately. Pick 1 to be put separately in 5 ways. The other 4 need to be split into 2 groups of 2 each. Pick 2 of the 4 to make the first group in 4C2 = 6 ways but since we have arranged this, we need to divide by 2! to get 3 ways.
So the 5 factors can be split into (2, 2, 1) in 5*3 = 15 ways. e.g. (6, 35, 11)

In all, 5 factors can be split into 3 distinct groups in 5 + 10 + 10 + 15 = 40 ways

KarishmaB · Answer

(1, 1, 2310) is not a valid case because sets have distinct elements.

RJ7X0DefiningMyX · Answer

In the second step, i.e. (multiple of 3 numbers)*(multiple of 1 number)*(multiple of 1 number) = 5c3*1*1=10, why did we not do  5C3*2C1*1C1  instead?

Saiganesh999 · Answer

Please check back that the order of a, b and c doesn't matter.
Here, a combination of (2*3*5, 7, 11) is same as (2*3*5, 11,7). So it is unnecessary to multiply 2C1 to the expression. 
This logic applies to all the combination questions as well. I hope its clear.

For how many three-element sets of positive integers {a, b, c} is it t

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For how many three-element sets of positive integers {a, b, c} is it t

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