Solution :

Required number of ways

= coefficient of x16 in (x3+x4+x5+.......x7)4

= coefficient of x16 in x12(1+x+x2+.......x4)4

= coefficient of x16 in x12(1−x5)4(1−x)−4

= coefficient of x4 in (1−x5)4(1−x)−4

= coefficient of x4 in (1−4x5+.....)

[1+4x+....+(r+1)(r+2)(r+3)3!x]

= (4+1)(4+2)(4+3)3! = 35

Aliter: Remaining 4 rupees can be distributed in 4+4−1C4−1 i.e., 35 ways

This can be mathematically formulated as

w+x+y+z=16 and each of them is greater or equal to 3 and we need to find number of all possible values of x w x y and z.

The number of non negative integral solution to the equation: (x1)+(x2)+(x3)+.....(xr)=n is given as (n+r-1)C(r-1)

Using above formula we will make a simple assumption

a=w-3, b=x-3, c=y-3 and d=z-3

So now the equation reduces

a+b+c+d=4

Solution is 7C3 that is 35