What is the number of seven digit integers, with sum of the digits equ

Question

What is 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?

(A) 55 
(B) 66
(C) 77 
(D) 88
(E) 99

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Foreheadson · Answer

the number can include only one 3, otherwise 3+3=6, even if all other 5 digits equal 1, sum is 11.
So if the number includes 3, it must also include exactly one 2 and all other digits should be 1. 
number of ways to write the number using this numbers is 7P2=42
the number may also invlude three 2s and four 1s. because 2s are the same we need 3C7=35
42+35=77, which is C.

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wishmasterdj · Answer

2 cases are possible:
1. 1111123 - 7!/5! = 42
2. 1111222 - 7!/4!*3! = 35

Hence 77   (C)

Alternatively, think in terms of Permutations - In such questions, since the sum constraint is provided, you ALREADY know WHICH digits to pick since the number of cases is delineated so, all you have to do is pick arrangements.

Hence 7P2 + 7P3 should cover both cases. 
Think about it

ScottTargetTestPrep · Answer

Solution:
 
If two of the seven digits were 3, then the sum of the digits would be greater than 10. We can verify this by assuming all the remaining five digits are as small as possible, i.e. 1. Thus, the seven-digit integer contains either no 3s or one 3. It follows that we could have only the following two options:

1) five 1s, one 2 and one 3

2) four 1s and three 2s

The first option has 7! / 5! = 7 x 6 = 42 integers, and the second option has 7! / (4! x 3!) = (7 x 6 x 5) / (3 x 2) = 35 integers. Therefore, we have a total of 42 + 35 = 77 such integers.

Answer: C

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What is the number of seven digit integers, with sum of the digits equ

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What is the number of seven digit integers, with sum of the digits equ

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