Math, asked by pshrimant2860, 1 year ago

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

Answers

Answered by amitnrw
10

Answer:

450

Step-by-step explanation:

3240 = 2 * 2 * 2 * 3 * 3 * 3 * 3 * 5

3240 = 2³ * 3⁴ * 5¹

3240 = a * b * c

Each integer Would have powers of 2 , 3 & 5

power for 2 can vary from 0 to 3  

power for 3 can vary from 0 to 4

power for 5 can vary from 0 to 1

The factor 2 can be split among 3 integers in 10 different ways:    (in terms of Power)

0,0,3

0,1,2

0,2,1

0,3,0

1,0,2

1,1,1

1,2,0

2,0,1

2,1,0

3,0,0

The factor 3 can be split among 3 integers in 15 different ways: (in terms of Power)

0,0,4

0,1,3

0,2,2

0,3,1

0,4,0

1,0,3

1,1,2

1,2,1

1,3,0

2,0,2

2,1,1

2,2,0

3,0,1

3,1,0

4,0,0

The factor 5 can be split among 3 integers in 3 different ways: (in terms of Power)

0,0,1

0,1,0

1,0,0

Total Combination = 10*15*3 = 450

Answered by vj7
3

Answer:

450

Step-by-step explanation:

The number 3240 can be factorized as 2 power 3

                                                                  3 power 4  

                                                                  5 power 1  

The formula for finding the number of ways

                                                         n-r-1 C r-1  

Applying for each term we get ,

                          2 power 3  ------> 3 + 3 - 1 C 3-1 ----> 5 C 2 = 10

                          3 power 4  ------> 4 + 3 -1  C 3-1 ----> 6 C 2 = 15

                          5 power 1  -------> 1 + 3 - 1 C 3-1 -----> 3 C 2 = 3  

The total number of ways = 10 * 15 * 3 = 450

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