Given:

The expression is (1!)¹ × (2!)² × (3!)³ × (4!)⁴ × … × (15!)¹⁵

Concept used:

To find the number of zeroes at the end of the product, we need to calculate the number of 2’s and number 5’s or number of pairs of 2 and 5.

2 × 5 = 10 ⇒ Number of zeroes = 1 (number of pair = 1)

The number of pairs of 2 and 5 is same as the number of zeroes at the end of the product

Calculation:

(1!)¹ × (2!)² × (3!)³ × (4!)⁴ × … × (15!)¹⁵

In the above expression, number of 2’s is more than number of 5’s. So we need to calculate the number of 5’s only

(1!)¹ × (2!)² × (3!)³ × (4!)⁴

There are so many 2’s and 5 is absent here, then number of zeroes = 0

(5!)⁵ = (5 × 4 × 3 × 2 × 1)⁵ ⇒ Number of 5’s = 5

(6!)⁶ = (6 × 5 × 4 × 3 × 2 × 1)⁶ ⇒ Number of 5’s = 6

Similarly, (7!)⁷ ⇒ Number of 5’s = 7

And up to (9!)⁹, number of 5’s is successively increased by 1

(10!)¹⁰ = (10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1)¹⁰ ⇒ Number of 5’s =  20

And up to (14!)¹⁴, number of 5’s is successively increased by 2

(15!)¹⁵ = (15 × 14 × 13 × 12 × 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1)¹⁵ ⇒ Number of 5’s = 45

Total number of 5’s = 5 + 6 + 7 + 8 + 9 + 20 + 22 + 24 + 26 + 28 + 45 = 200

Total number of pairs of 2 and 5 = 200