Question

In how many ways can a set of two nonnegative integers less than 100 be chosen?

Answer 1

Answer: 4950

Step-by-step explanation:

The number of possible combinations of n things taken r at a time is given by :-

$C(n,r)=\dfrac{n!}{r!(n-r)!}$

Total nonnegative integers less than 100 ={0,1,2,... ,99} = 100

So, the number of combinations of choosing 2 out of them = $C(100,2)=\dfrac{100!}{2!98!}=\dfrac{100\times99\times98!}{2\times98!}\\\\=\dfrac{100\times99}{2}\\\\=50\times 99\\\\=4950$

So, the number of ways to choose a set of two nonnegative integers less than 100 = 4950