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Q.
The number of three elements sets of positive integers $\{a, b, c\}$ such that $a \times b \times c=2310,$ is
Permutations and Combinations
Solution:
$N=2310=2 \cdot 3 \cdot 5 \cdot 7 \cdot 11$
We need to either partition these $5$ prime numbers in groups of $2$ with the other third number $1$ or groups of $3$ .
Now
$5=1+4=2+3 $ and $5=1+1+3=1+2+2$
Hence, number of ways
$=\frac{5 !}{1 ! 4 !}+\frac{5 !}{2 ! 3 !}+\frac{5 !}{1 ! 1 ! 2 ! 3 !}+\frac{5 !}{1 ! 2 ! 2 ! 2 !} $
$=5+10+10+15=40$