Permutation and combination

Question

Find the no. of seven digit integers with sum of digits = 11 and formed by using the digits 1,2 and 3 only ?

Comments

here summation is 11, a small number, we can partition 11 into 3 groups. But what about large digit sum ? should we use this partition method or use generating function ?

Answer 1

7 digit number by 1,2,3 and sum 11

The number will be combination of 2 2 2 2 1 1 1 =$\frac{7!}{4!3!}$=35

or, 3 3 1 1 1 1 1=$\frac{7!}{2!5!}$=21

or, 3 2 2 1 1 1 1=$\frac{7!}{2!4!}$=105

So, total such numbers possible=35+21+105=161

Comments

First of all it depends on question and I don't think in gate exam we will be asked such question tough question ..

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question is not trying, its about time. take any digit sum like 48 (say) suing 1,2,3,4

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@Debashish. I think you have solved one recurrance like this. The recurrance would go like this:

let, P(sum) = #ways/permutations to form the given sum

Thus, P(sum) =

P(sum - 3)   (if 3 is the MSB)

+ P(sum - 2) (if 2 is the MSB)

+ P(sum - 1) (if 1 is the MSB)

Also, questions in GATE are such that they can be solved in 1 or 2 minute(s) mostly with simple techniques unless tricky :)