JEE Main 2025 Jan 29 Shift 1, Mathematics Q6: Combinations

JEE Main2025Jan 29, Shift 1Mathematics

Let $P$ be the set of seven digit numbers with sum of their digits equal to $11$ . If the numbers in $P$ are formed by using the digits $1, 2$ and $3$ only, then the number of elements in the set $P$ is :

A
$158$
B
$173$
C
$164$
D
$161$

Solution

Let the seven digits comprise $a$ ones, $b$ twos and $c$ threes. Then $a + b + c = 7$ and $a + 2 b + 3 c = 11$ , which gives $b + 2 c = 4$ with $a, b, c \geq 0.$

Non-negative integer solutions $(a, b, c)$ : $(3, 4, 0), (4, 2, 1), (5, 0, 2) .$

Number of distinct arrangements for each multiset:

$(3, 4, 0)$ : digits ${1, 1, 1, 2, 2, 2, 2}$ , count $= \frac{7 !}{3 ! 4 !} = 35.$

$(4, 2, 1)$ : digits ${1, 1, 1, 1, 2, 2, 3}$ , count $= \frac{7 !}{4 ! 2 ! 1 !} = 105.$

$(5, 0, 2)$ : digits ${1, 1, 1, 1, 1, 3, 3}$ , count $= \frac{7 !}{5 ! 2 !} = 21.$

Total $= 35 + 105 + 21 = 161.$

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JEE Main 2025 Jan 29 Shift 1, Mathematics Q6: Combinations

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