JEE Mathematics 2023 Question with Solution

Given: Digits are $1,2,2,2,3,3,5$.

Find: The number of seven-digit odd numbers that can be formed using all these digits.

For the number to be odd, the unit digit must be $1$, $3$, or $5$.

If unit digit is $5$, then the remaining digits are arranged in $\frac{6!}{3!2!}$ ways.

$$\frac{6!}{3!2!}=60$$

If unit digit is $3$, then the remaining digits are $1,2,2,2,3,5$, which can be arranged in $\frac{6!}{3!}$ ways.

$$\frac{6!}{3!}=120$$

If unit digit is $1$, then the remaining digits are arranged in $\frac{6!}{3!2!}$ ways.

$$\frac{6!}{3!2!}=60$$

Therefore, total numbers are

$$60+60+120=240$$

Therefore, the required number is $240$.

Given: Digits are $1,2,2,2,3,3,5$.

Find: Total number of seven-digit odd numbers.

1. An odd number must end in an odd digit.
2. Available odd digits are $1,3,5$.
3. Count arrangements for each possible last digit separately.

If the last digit is $5$, the first six places contain $1,2,2,2,3,3$.

$$\text{Number of arrangements} = \frac{6!}{3!2!}=60$$

If the last digit is $3$, the first six places contain $1,2,2,2,3,5$.

$$\text{Number of arrangements} = \frac{6!}{3!}=120$$

If the last digit is $1$, the first six places contain $2,2,2,3,3,5$.

$$\text{Number of arrangements} = \frac{6!}{3!2!}=60$$

Adding all cases,

$$60+120+60=240$$

So, the correct answer is $240$.