Permutation and Combination – Permutations with Repetition – JEE Main 29 Jan 2025 Shift 1

Solution:

We need to form 7-digit numbers using digits 1, 2, 3 such that the sum of digits is 11.

Let the number of 1s be $n_1$, 2s be $n_2$, and 3s be $n_3$.

Total digits: $n_1 + n_2 + n_3 = 7$.

Sum of digits: $1(n_1) + 2(n_2) + 3(n_3) = 11$.

Subtracting the first equation from the second:

$(n_1 + 2n_2 + 3n_3) – (n_1 + n_2 + n_3) = 11 – 7$

$n_2 + 2n_3 = 4$.

Possible cases for $(n_3, n_2)$:

**Case 1:** $n_3 = 2$

$n_2 + 2(2) = 4 \Rightarrow n_2 = 0$.

$n_1 = 7 – 0 – 2 = 5$.

Digits: {1, 1, 1, 1, 1, 3, 3}.

Number of ways = $\frac{7!}{5!2!} = \frac{7 \times 6}{2} = 21$.

**Case 2:** $n_3 = 1$

$n_2 + 2(1) = 4 \Rightarrow n_2 = 2$.

$n_1 = 7 – 2 – 1 = 4$.

Digits: {1, 1, 1, 1, 2, 2, 3}.

Number of ways = $\frac{7!}{4!2!1!} = \frac{7 \times 6 \times 5}{2} = 105$.

**Case 3:** $n_3 = 0$

$n_2 + 2(0) = 4 \Rightarrow n_2 = 4$.

$n_1 = 7 – 4 – 0 = 3$.

Digits: {1, 1, 1, 2, 2, 2, 2}.

Number of ways = $\frac{7!}{3!4!} = \frac{7 \times 6 \times 5}{6} = 35$.

Total number of elements in $P = 21 + 105 + 35 = 161$.

Ans. (4)