Question

In how many ways can we write 20 as a sum of 3 non-negative integer?

Answer 1

Answer:

Step-by-step explanation:

This is a question on permutation.

With permutation the order of the arrangements is necessary.

The formula for permutation is as follows :

Given n objects from which you need to choose r objects at a time the formula for the permutation is as follows :

p = $\frac{n!}{(n - r)!}$

From the question above we have :

n = 20

r = 3

We need to get the numbers that when picked three at a time can sum up to 20.

We do the substitution as follows :

P = 20!/(20-3)!

This equals to :

20!/17! = 6840

The answer is 6840 ways.

Answer 2

Answer:

Step-by-step explanation:

Let x, y, and z be 3 non negative integer.

To find: No ways such that 20 = x + y + z

we find this using combination of numbers for total of n+r-1 items out of which n would be same (of 1 type) and the other r - 1 would be same as well (of another type)

i.e.,we use,

$^{n+r-1}\textrm{C}_{r-1}$

$\implies\frac{(n+r-1)!}{(r-1)!\times(n+r-1-(r-1))!}$

$\implies\frac{(n+r-1)!}{(r-1)!\timesn!}$

here n = 20 & r = 3

So, the no. of ways is given by

$^{20+3-1}\textrm{C}_{3-1}\:=\:^{22}\textrm{C}_{2}$

$\implies\frac{22!}{2!\times20!}$

$\implies\frac{22\times21\times20!}{2\times20!}$

$\implies\frac{22\times21}{2}$

$\implies11\times21$

$231$

Therefore, No. of ways in which 20 can be writtenas sum of 3 non negative interger is 231.