The number of seven-digit integers, with sum of the digits equal to 10 and formed by using the digits 1, 2 and 3 only, is
- A.55
- B.66
- C.77
- D.88
Show answer and solution
Correct option: C
Solution
We need 7-digit numbers using digits 1, 2, 3 only, with digit sum = 10.
Let a, b, c = count of 1s, 2s, 3s respectively.
Then: a + b + c = 7 and a + 2b + 3c = 10, giving b + 2c = 3.
Cases for (b, c): (3,0), (1,1).
Case 1: c=0, b=3, a=4 → digits: four 1s, three 2s, zero 3s
Arrangements = 7!/(4!3!) = 35
Case 2: c=1, b=1, a=5 → digits: five 1s, one 2, one 3
Arrangements = 7!/(5!1!1!) = 42
Total = 35 + 42 = 77
Let a, b, c = count of 1s, 2s, 3s respectively.
Then: a + b + c = 7 and a + 2b + 3c = 10, giving b + 2c = 3.
Cases for (b, c): (3,0), (1,1).
Case 1: c=0, b=3, a=4 → digits: four 1s, three 2s, zero 3s
Arrangements = 7!/(4!3!) = 35
Case 2: c=1, b=1, a=5 → digits: five 1s, one 2, one 3
Arrangements = 7!/(5!1!1!) = 42
Total = 35 + 42 = 77
[AIEEE 2009]