MHB Factoring $27720$ into Co-Prime Factors

[juantheron]

The number of ways in which the number $27720$ can be split into two factors which are co prime

 

[kaliprasad]

The number of ways in which the number $27720$ can be split into two factors which are co prime

Spoiler

we have $27720 = 2^3 * 3^2 *5 * 7 * 11$
the number of factors which are coprimes 2^5 = 32
so it can be factored is 16 out of which we should leave out 1 * 27720

so number of ways = 15

 

[mathbalarka]

Spoiler

$$27720 = 2^3 \cdot 3^2 \cdot 5 \cdot 7 \cdot 11$$

Note that multiple factors doesn't really play a part, as we have required the two parts to be coprime. Thus, it's essentially equivalent to write out the the number of ways the *square-free part of $27720$*, i.e., $n = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11$ can be written as a product of two coprime factors.

But this in turn is equivalent to partition the set $\{2, 3, 5, 7, 11\}$ into two disjoint sets, as all of the factors of $n$ are relatively coprime.

Then that again is equivalent to partition $5$ in two nonzero parts, not necessarily ordered. Enumerating gives :

$$\begin{aligned}5 \;&= 4 + 1 = 1 + 4 \\ &= 2 + 3 = 3 + 2 \\ &= 3 + 1 + 1 = 1 + 3 + 1 = 3 + 1 + 1 \\ &= 1 + 2 + 2 = 2 + 1 + 2 = 2 + 2 + 1 \\ &= 2 + 1 + 1 + 1 = 1 + 2 + 1 + 1 = 1 + 1 + 2 + 1 = 1 + 1 + 1 + 2 \\ &= 1 + 1 + 1 + 1 + 1 \end{aligned}$$

Which is a total of $2 + 2 + 3 + 3 + 4 + 1 = 15$ partitions. Thus there are $15$ ways to express $27720$ as a product of two coprime numbers $\blacksquare$

 

[juantheron]

Thanks kaliprasad,mathbalarka(For Nice explanation.)

My solution is same as kaliprasad.

 

[Albert1]

the number of ways in which the number $27720$ can be split into two factors which are co prime

[sp]
$\dfrac{C_1^5 +C_2^5+C_3^5+C_4^5}{2}=15$
or :

$C_1^5 +C_2^5=15$
[/sp]