Question

In how many possible ways can write 3240 as a product of 3 positive integers a,b and c.

Answer 1

$3240$ = $2^{3}$ × $3^{4}$ × $5$

$|||^{ly}$ for 3 and 5 

∴ total = 10 × 15 × 3 = 450

Hope I helped you! :)

Answer 2

Answer:

Number of ways is 450.

Step-by-step explanation:

Given:

Number is 3240

To find: Number of ways in which given number expressed as product of three positive integers.

Prime factorization of 3240 = $2^3\times3^4\times5^1$

So,

Number of ways in which 3 2's can be distributed to 3 integers a , b and c = $^{3+(3-1)}C_{3-1}=^5C_2=\frac{5!}{2!(5-2)!}=\frac{5\times4\times3!}{2\times3!}=10$

Number of ways in which 4 3's can be distributed to 3 integers a , b and c = $^{4+(3-1)}C_{3-1}=^6C_2==\frac{6!}{2!(6-2)!}\frac{6\times5\times4!}{2\times4!}=15$

Number of ways in which 1 5's can be distributed to 3 integers a , b and c = $^{1+(3-1)}C_{3-1}=^3C_2=3$

Therefore, total number of ways = 10 × 15 × 3 = 450 ways.

Therefore, Number of ways is 450.