Here, the maximum value that m can take is n and same for k.
Tried with examples but m not able to come to a general solution.
When m=k, the no. of ways =1
When m>k, the solution is
(m+1)C{k+1) + mC(k+1) + (m-1)C(k+1)…
A stupid solution I know. Working on generalizing this.
Still in CAT the best way to tackle such questions would be to go for value of n=5/6 and the values of k upto 5 and similarly for m.

Here, the maximum value that m can take is n and same for k.

Tried with examples but m not able to come to a general solution.

When m=k, the no. of ways =1

When m>k, the solution is

(m+1)C{k+1) + mC(k+1) + (m-1)C(k+1)…

A stupid solution I know. Working on generalizing this.

Still in CAT the best way to tackle such questions would be to go for value of n=5/6 and the values of k upto 5 and similarly for m.

milindOctober 20, 2008 at 4:42 pm

its simpler

the highest term can be selected in one way

as it is m+1

now other k terms have to be selected from m values

so mck

outtimedOctober 21, 2008 at 12:36 pm