Question

The number of ways in which we can select three numbers from $1$ to $30$ so as to exclude every selection of all even numbers is?

• A. $4060$
• B. $3605$
• C. $455$
• D. None of these

Answer 1

The correct option is B

$3605$

Explanation for the correct option(s)

Find the number of ways.

We need to select three numbers from $1$ to $30$ such that selection of all even numbers is excluded.

We know that

${}^n{C}_r=\frac{n!}{r!\left(n-r\right)!}$

Total number of ways can be expressed as,

$$\begin{gathered} {}^{30}{C}_3=\frac{30!}{3!\left(30-3\right)!} \\ {}^{30}{C}_3=\frac{30!}{3!\times 27!} \\ {}^{30}{C}_3=\frac{30\times 29\times 28\times 27!}{3\times 2\times 1\times 27!} \\ {}^{30}{C}_3=10\times 29\times 14 \\ {}^{30}{C}_3=4060 \end{gathered}$$

Number of ways of selecting even numbers can be expressed as,

$$\begin{gathered} {}^{15}{C}_3=\frac{15!}{3!\left(15-3\right)!} \\ {}^{15}{C}_3=\frac{30!}{3!\times 12!} \\ {}^{15}{C}_3=\frac{15\times 14\times 13\times 12!}{3\times 2\times 1\times 12!} \\ {}^{15}{C}_3=5\times 7\times 13 \\ {}^{15}{C}_3=455 \end{gathered}$$

Thus, the number of ways when all even numbers are excluded can be expressed as,

$4060-455=3605$

Hence, the correct answer is Option B.