How will you do it ? (III)

Computer Science Level 3

$\large \sum_{a=1}^{10} \sum_{b=a}^{10} \sum_{c=b}^{10}{\left( {a \times b+c} \right)}$

What is the number of terms in the series given above?

Details and Assumptions:

$\left({1 \times 1+1}\right)$ and $\left({1 \times 1+2}\right)$ are the $1^{\textrm{st}}$ and the $2^{\textrm{nd}}$ terms of the series respectively.

this problem is a part of the set Fundamental Programming

220 202 101 110

4 solutions

Arulx Z
Jan 5, 2016

Python one-liner

1	`sum(1 for a in xrange(1, 11) for b in xrange(a, 11) for c in xrange(b, 11))`

Masbahul Islam
Jan 4, 2016

Excel vba code:

s = 0

For a = 1 To 10

For b = a To 10

For c = b To 10

s = s + 1

Next c, b, a

MsgBox s

Zeeshan Ali
Jan 3, 2016

Here is an algorithm which does your required work;

    terms<-0
    for a<-1 to 10
        for b<-a to 10
            for c<-b to 10
                terms<-terms+1
    return terms

Isn't that so simple.!?

Moderator note:

What is the combinatorics approach to this? In particular, note that ${ 12 \choose 3 } = 220$ .

I don't think it can be generalized in this way for any integer n such that the series is: $\large \sum_{a=1}^{n} \sum_{b=a}^{n} \sum_{c=b}^{n}{\left( {a \times b+c} \right)}=\left({1 \times 1+1}\right)+\left({1 \times 1+2}\right)+...$ Moreover I am NOT requiring the sum rather I wanna know the total number of terms of that series.

Zeeshan Ali - 5 years, 5 months ago

Algebraic: Basically, with a=1, c is from 1-10, plus 2-10, plus 3-10, etc... is 10+9+8...+1 terms with a=2, c is from 2-10, 3-10, 4-10, etc... --> is 9+8+7...+1 terms 1 ten, 2 nines, 3 eights, 4 sevens, 5 sixes. You can go all the way to ten or multiply that by 2 since the equation will be symmetrical. (10+9 2+8 3+7 4+6 5)*2

Alex Li - 5 years, 1 month ago

We can show that C(12,3) is the answer. For each triplet (a,b,1,c1) from C(12,3) we take triplet (a,b1-1,c1-2) and b1-1>=a, c1-2>=b so it is part of our sum.

Hasmik Garyaka - 3 years, 8 months ago

Hasmik Garyaka
Oct 15, 2017

(n+1)3-n3=3n2+3n+1

summa((0,n)3n2+3n+1)=(n+1)3

summa((0,n)3n2+3n)=(n+1)3-(n+1)=n(n+1)(n+2)

How will you do it ? (III)

this problem is a part of the set Fundamental Programming

4 solutions

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