P & C – Fundamental Principle of Counting – JEE Main 28 January 2025 Shift 1

Question ID: #426
JEE Main28 January Shift 1, 2025Algebra

The number of different 5 digit numbers greater than 50000 that can be formed using the digits 0, 1, 2, 3, 4, 5, 6, 7, such that the sum of their first and last digits should not be more than 8, is

  • (1) 4608
  • (2) 5720
  • (3) 5719
  • (4) 4607

Solution:


Let the 5-digit number be $d_1 d_2 d_3 d_4 d_5$.

Given digits are $\{0, 1, 2, 3, 4, 5, 6, 7\}$ (Total 8 digits).

Since the number must be greater than $50000$, the first digit $d_1$ can be $5, 6,$ or $7$.

The condition given is $d_1 + d_5 \le 8$.

Case 1: If $d_1 = 5$

$5 + d_5 \le 8 \Rightarrow d_5 \le 3$. So, $d_5 \in \{0, 1, 2, 3\}$ (4 possibilities).

The digits $d_2, d_3, d_4$ can be any of the 8 digits.

Total numbers = $1 \times 8 \times 8 \times 8 \times 4 = 2048$

Case 2: If $d_1 = 6$

$6 + d_5 \le 8 \Rightarrow d_5 \le 2$. So, $d_5 \in \{0, 1, 2\}$ (3 possibilities).

Total numbers = $1 \times 8 \times 8 \times 8 \times 3 = 1536$

Case 3: If $d_1 = 7$

$7 + d_5 \le 8 \Rightarrow d_5 \le 1$. So, $d_5 \in \{0, 1\}$ (2 possibilities).

Total numbers = $1 \times 8 \times 8 \times 8 \times 2 = 1024$

Total ways = $2048 + 1536 + 1024 = 4608$

However, we need numbers strictly greater than $50000$.

The case where $d_1=5$ and $d_2=d_3=d_4=d_5=0$ forms the number $50000$ exactly. Since $0 \in \{0, 1, 2, 3\}$, this number is included in Case 1.

We must subtract this single case.

Required numbers = $4608 – 1 = 4607$

Ans. (4)

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