JEE Main 2025 — 29 January, Shift 1. Previous Year Question.
Problem
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:
Key insight. With only three digits allowed and seven positions to fill, there are just a handful of digit combinations whose sum is 11 — the real work is finding all of them systematically (not by guessing), then counting the arrangements of each combination separately and adding them up.
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Approach
Since every digit is either 1, 2 or 3, and there are exactly 7 digits, start from “all seven digits are 1” (which sums to 7) and work out how many of those 1s need to be swapped for 2s or 3s to push the total up to 11. Each valid swap-combination is a separate case, and each case needs its own permutation count, since it involves a different multiset of repeated digits.
Solution
Step 1 — Find every digit combination that sums to 11
Starting from seven 1s (sum $=7$), the shortfall to reach 11 is $4$, which has to be made up by replacing some 1s with 2s and/or 3s. Working through the possibilities systematically:
- Five 1s, two 3s: $5(1) + 2(3) = 5+6 = 11$ ✓
- Four 1s, two 2s, one 3: $4(1)+2(2)+1(3) = 4+4+3 = 11$ ✓
- Three 1s, four 2s: $3(1)+4(2) = 3+8 = 11$ ✓
Checking further (e.g. using only two 1s) pushes the required sum from the remaining five digits above what 2s and 3s can reach without overshooting or needing a fourth combination — no other cases work.
Step 2 — Count the arrangements of each case
Each case is a multiset of 7 digits with repeats, so the count of distinct 7-digit numbers is $\dfrac{7!}{(\text{repeats})!}$.
Case 1 (five 1s, two 3s):
$$\frac{7!}{5!\,2!} = \frac{7\times 6}{2} = 21$$
Case 2 (four 1s, two 2s, one 3):
$$\frac{7!}{4!\,2!\,1!} = \frac{7\times6\times5}{2} = 105$$
Case 3 (three 1s, four 2s):
$$\frac{7!}{3!\,4!} = \frac{7\times6\times5}{3\times2\times1} = 35$$
Step 3 — Add the cases together
$$21 + 105 + 35 = 161$$
Answer
$$161$$
Common mistakes
- Missing a valid case, or double-counting an invalid one. Since the case-search is done by hand, it’s worth double-checking each combination’s digit sum before computing its permutation count — an error here throws off the whole answer.
- Forgetting to divide by the factorial of every repeated digit. In Case 2 there are two different repeats (four 1s and two 2s), and both factorials — $4!$ and $2!$ — belong in the denominator, not just one.
Practise next
- Find the number of seven-digit numbers using only the digits 1, 2 and 3 whose digits sum to 12, using the same case-by-case method.
Show answer
$266$. Let the seven digits contain $p$ ones, $q$ twos and $r$ threes, so $p+q+r=7$ and $p+2q+3r=12$, giving $q+2r=5$.
$r=0$: $q=5$, $p=2$ — $\dfrac{7!}{2!\,5!}=21$. $r=1$: $q=3$, $p=3$ — $\dfrac{7!}{3!\,3!\,1!}=140$. $r=2$: $q=1$, $p=4$ — $\dfrac{7!}{4!\,1!\,2!}=105$.
Total $21+140+105=266$.
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