UPSC NDA I 2025 Mathematics Paper

Total number of items, n=6
We need to calculate the total number of selections for r=0,1,2, and 3:
For r=0
‌6C0=‌

6!

0!(6−0)!

=‌

1

1

=1
⇒ Number of ways to select 0 items =1
For r=1
‌6C1=‌

6!

1!(6−1)!

=‌

6

1

=6
⇒ Number of ways to select 1 item =6
For r=2
‌6C2=‌

6!

2!(6−2)!

=‌

6×5

2×1

=15
⇒ Number of ways to select 2 items =15
For r=3
‌6C3=‌

6!

3!(6−3)!

=‌

6×5×4

3×2×1

=20
⇒ Number of ways to select 3 items = 20

Total number of selections:
‌6C0+‌6C1+‌6C2+‌6C3=1+6+15+20=42
∴ The total number of selections of at most 3 things from 6 different things is 42 .