$\frac{n*(n+1)*(n+2)*(n+3)*(n+4)*(n+5)}{1.2.3.4.5.6}$

is this the answer ?

is this the answer ?

2 votes

void foo(int n) { for(i1=1;i1<=n;i1++) { for(i2=1;i2<=i1;i2++) { ....... { for(i6=1;i6<=i5;i6++) { count++; } } } } }

Count initially 0.What is value returned by foo(8)?

5

$S_1=n$ , if there was only $1$ for loop

$S_2=\sum_{1}^{n} n=\frac{n*(n+1)}{1*2}$ , if there were $2$ for loop

$S_3=\sum_{1}^{n}S_2=\frac{n*(n+1)*(n+2)}{1*2*3}$ , if there were $3$ for loop

similarly, $S_6=\sum_{1}^{n}S_5=\frac{n*(n+1)*(n+2)*(n+3)*(n+4)*(n+5)}{1*2*3*4*5*6}$ , if there were $6$ for loop

That's how i did it

$S_2=\sum_{1}^{n} n=\frac{n*(n+1)}{1*2}$ , if there were $2$ for loop

$S_3=\sum_{1}^{n}S_2=\frac{n*(n+1)*(n+2)}{1*2*3}$ , if there were $3$ for loop

similarly, $S_6=\sum_{1}^{n}S_5=\frac{n*(n+1)*(n+2)*(n+3)*(n+4)*(n+5)}{1*2*3*4*5*6}$ , if there were $6$ for loop

That's how i did it

0

@Shobhit Joshi's answer is nice.

I just want to add that, this type of loop is a general form for computing (n+r-1)C(r), here n=8, r=6.