Find the no.of 4 - digit no.s that can be formed using the digits 1,2,5,6,7. How many of them are divisible by (i) 2 (ii) 3 (iii) 4 (iv) 5 (v) 25.

The no.of 4 digited no. that can be formed using the digits 1,2,5,6,7 is ⁵P₄ = 120

i) A no. is divided by ‘2’ when its unit place must be filled with an even digit from among the given integers. This can be done in 2 ways. Now, the remaining 3 places can be filled with 4 digits is⁴P₃ = 24
∴ total no. of ways = 2 × 24 = 48.

ii) A no. is divisible by 3 only when the sum of the digits in that no is a multiple of 3. Sum of the Given 5 digits 1+2+5+6+7 = 21. The 4 digits such that, their sum is multiple of 3 from the given digits are 1,2,5,7. They can be arranged in 4 ways and all these 4 digited no.s are divisible by 3. The no. of 4 digited no.s divisible by 3 = 4! = 24.

iii) A no. is divisible by 4 only when the last two places of it is multiple of 4. Hence, the last two places should be filled with
one of the following 12, 16, 52, 56, 72, 76. Thus the last two places can be filled in 6 ways. The remaining 3 digits in ³P₂ = 6 ways. 

∴ The no. of 4 digits no.s divisible by 4 = 6 × 6 = 36.

iv) A no. is divisible by 5 when its units place must filled by 5 from the given integers 1, 2, 5, 6, 7. This can be done in one way
The remaining 3 places can be filled with remaining 4 digits in ⁴P₃ = 24 ways. 

∴ The no. of 4 digited no.s divisible by ‘5’  = 1 × 24 = 24.

v) A no. is divisible by 25 when its last two places are filled with either 25 (or) 75. Thus the last two places can be filled in 2 ways. the remaining 2 places from the remaining 3 digits.can be filled in ³P₂ = 6 ways. 

∴ The no. of 4 digited no.s divisible by 25 = 2 × 6 = 12