Question

In how many ways can 3 integers be selected from the set {1, 2, 3, ….., 37} such that the sum of the 3 integers is an odd number

Answer 1

Answer: 3876

Here, even + odd+odd =odd

Odd+odd +odd=odd

So.,from 1 to 37 there are 19 odd numbers and 18 even numbers so either 3 odd numbers from 19 can be selected or 2 even numbers from 18 and odd number from 19 can be selected.

Hence, 19C3 +( 19C1 x 18C2) =3876

Answer 2

We need to recall the following rules.

• Odd $+$ Odd $+$ Odd $=$ Odd
• Even $+$ Odd $+$ Even $=$ Odd
• $^nC_r=\frac{n!}{r!(n-r)!}$

Given:

$S=\{1,2,3,....,37\}$

To Find: The number of ways of selecting $3$ integers such that the sum of the $3$ integers is an odd number.

In the given set $S$,

number of odd integers $=19$

number of even integers $=18$

Case $1$: All three integers are odd

The number of ways will be,

$^{19}C_3=\frac{19!}{3!(19-3)!}$

$^{19}C_3=\frac{19!}{3!(16)!}$

$^{19}C_3=\frac{19\times 18\times 17}{6}$

$^{19}C_3=969$

Case $2$: Two integers are even and one integer is odd

The number of ways will be,

$19\times ^{18}C_2=19\times \frac{18!}{2!(18-2)!}$

$19\times ^{18}C_2=19\times \frac{18!}{2!(16)!}$

$19\times ^{18}C_2=19\times \frac{18\times 17}{2}$

$19\times ^{18}C_2=2907$

Hence, the total number of ways of selecting $3$ integers such that the sum of the $3$ integers is an odd number is,

Total ways $=^{19}C_3+19\times ^{18}C_2$

Total ways $=969+2907$

Total ways $=3876$