First slide

Combinations

Question

If $10! = 2^{p} \cdot 3^{q} \cdot 5^{r} \cdot 7^{s}$ , then

Moderate

A

2q = p
B

pqrs = 64
C

number of divisors of 10! is 280
D

number of ways of putting 10! as a product of two natural numbers is 135

Solution

Exponent of 2 is
$[\frac{10}{2}] + [\frac{10}{2^{2}}] + [\frac{10}{2^{3}}] = 5 + 2 + 1 = 8$
Exponent of 3 is
$[\frac{10}{3}] + [\frac{10}{3^{2}}] = 3 + 1 = 4$
Exponent of 5 is
$[\frac{10}{5}] = 2$
Exponent of 7 is
$[\frac{10}{7}] = 1$
The number of divisors of 10! is (8 + 1) (4 + 1) (2 + 1) (1 + 1) = 270. The number of ways of putting N as a product of two natural numbers is 270/2 = 135.

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