JEE Main 2025 Jan 28 Shift 1, Mathematics Q1: Permutation

JEE Main2025Jan 28, Shift 1Mathematics

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

A
$4608$
B
$5720$
C
$5719$
D
$4607$

Solution

The first digit must be one of ${5, 6, 7}$ to make the number exceed $50000$ , and digits are drawn from the set ${0, 1, \dots, 7}$ with repetition allowed.

For each leading digit $d_{1}$ , the last digit $d_{5}$ must satisfy $d_{1} + d_{5} \leq 8$ . Count permissible last digits:

$d_{1} = 5 \Rightarrow d_{5} \in {0, 1, 2, 3}$ , 4 choices.
$d_{1} = 6 \Rightarrow d_{5} \in {0, 1, 2}$ , 3 choices.
$d_{1} = 7 \Rightarrow d_{5} \in {0, 1}$ , 2 choices.

Total $(d_{1}, d_{5})$ pairs $= 4 + 3 + 2 = 9$ . The middle three digits are free over the 8-digit set: $8^{3} = 512$ arrangements each.

Total $= 9 \times 512 = 4608$ . But the count includes $50000$ itself, which is not greater than $50000$ . Subtract it.

$∴$ Answer $= 4608 - 1 = 4607$ .

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JEE Main 2025 Jan 28 Shift 1, Mathematics Q1: Permutation

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