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Clearly the factors are $2310=2\times 3\times 5 \times 7 \times 11$

My approach was picking one number for $a$, this leaves 4 numbers for b and c. So b can take up 1 or 2 or 3 numbers out of those 4 numbers

So $$\binom 51 (\binom 41 +\binom 42 +\binom 43)$$ $$=5(2^4-2)$$

Similar, 2 numbers can be picked for $a$, leaving 3 for b and c

Continuing this way gives $$5(2^4-1) +10 (2^3-2) + 10 (2^2-2) $$which isn’t even close to 40, the answer.

What am I doing wrong?

I know this question already exists on MSE, but unfortunately it doesn’t solve my specific problem.

NOTE: I am using this problem as a reference for my answer The students $S_1, S_2,...S_{10}$ are divided into 3 groups A, B and C

Aditya
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1 Answers1

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Knowing the prime factors makes things a lot easier. We basically just assign each factor to any of $a$, $b$, or $c$ and call it good. This means that there are a total of $3^5$ possible ways of assigning these factors, a total of $243$ ways.

This, however, overcounts several conditions. Firstly, if we ever end up in a situation where two of the numbers are the same, then we don't get a $3$ element set, so we need to exclude these. These conditions occur if ever the distribution of the prime factors could lead to two equal products. But, by the uniqueness of prime factorization, since all of our primes are distinct this can only happen if two of the numbers were both $1$, meaning that all of the factors ended up in a single number. This happens in exactly $3$ situations, so we remove these three.

We have still overcounted, however, since we do not care about the permutations of $a$, $b$, and $c$. We divide this new total by the total number of permutations that would lead to this collection of numbers, which is a total of $3!$ permutations. In all, we have

$$\frac{3^5-3}{3!}=\frac{243-3}{6}=\frac{240}{6}=40$$

Hope this helps!

Josh B.
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  • Just confirming here, the sets you eliminated were {1,x,y},{1,1,x} and one I am missing – Aditya Mar 08 '21 at 17:55
  • The actual answer is 41 -- you have subtracted out the three permutations of $(1, 1, 2310)$ to consider only those cases where you have three distinct values, but have forgotten to add it back in. – Nick Peterson Mar 08 '21 at 17:59
  • @NickPeterson No, the answer is $40$. – jjagmath Mar 08 '21 at 18:01
  • @jjagmath I can list them, if you like... – Nick Peterson Mar 08 '21 at 18:06
  • @jjagmath $(1, 1, 2310), (1, 2, 1155), (1, 3, 770), (1, 5, 462), (1, 6, 385), (1, 7, 330), (1, 10, 231), (1, 11, 210), (1, 14, 165), (1, 15, 154), (1, 21, 110), (1, 22, 105), (1, 30, 77), (1, 33, 70), (1, 35, 66), (1, 42, 55), (2, 3, 385), (2, 5, 231), (2, 7, 165), (2, 11, 105), (2, 15, 77), (2, 21, 55), (2, 33, 35), (3, 5, 154), (3, 7, 110), (3, 10, 77), (3, 11, 70), (3, 14, 55), (3, 22, 35), (5, 6, 77), (5, 7, 66), (5, 11, 42), (5, 14, 33), (5, 21, 22), (6, 7, 55), (6, 11, 35), (7, 10, 33), (7, 11, 30), (7, 15, 22), (10, 11, 21), (11, 14, 15)$ – Nick Peterson Mar 08 '21 at 18:07
  • @NickPeterson Are you including {1,1,2310}? Becuase {1,1,2310} is not a 3 element set. – jjagmath Mar 08 '21 at 18:08
  • Ah-hah. Yes, right you are. My bad. :-) – Nick Peterson Mar 08 '21 at 18:08
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    @Aditya The first three eliminated sets are ${2310,1,1},{1,2310,1}$, and ${1,1,2310}$, since these are not 3 element sets. – Josh B. Mar 08 '21 at 21:57