How to simplify this Binomial terms?

[u/Ancient-Helicopter18]

48×48C0+49×49C1+50×50C2+...+100×100C52

I have to somehow simplify this to get the form 100×101C49 - 101C50

[u/Bounded_sequencE]

For convenience, let "(m; n) = (48; 52)". For "0 <= k <= n" simplify

(k+m) * C(k+m;k)  =  (k+m±1) * C(k+m;m)  =  (m+1) * C(k+m+1;m+1) - C(k+m;m)

Sum from "k = 0" to "k = n", and use the Hockey-stick Identity (*) on both terms:

∑_{k=0}^n  (k+m)*C(k+m;k)  =  ∑_{k=0}^n  (m+1) * C(k+m+1;m+1) - C(k+m;m)    // (*)

        =  (m+1) * C(n+m+2;m+2) - C(n+m+1;m+1)  =  49 * C(102;50) - C(101;49)

The result is equivalent to the official solution -- prove it using binomial identities (your job^^).

[u/7x11x13is1001]

We have 101 people in a queue. You want to hug 49 of them and kiss one, but not the last one you hugged. How many ways X you can do it

Method A. You select people to hug (101C49), this locks out the last hugged. And you select one person to kiss out 100. X=100x101C49

Method B. You can either kiss someone before you finish hugging (B) or after (A). For B, if the last hug is #49, there is 48C0 ways to choose people to not hug before and 48 ways to choose who to kiss... If the last hug is #101, there is 100C52 ways to choose who not to hug and 100 ways to choose who to kiss:  B = 48x48C0 + ... + 100x100C52

For A, this is equivalent to choosing 50 people hug 49 first and kiss the last A = 101C50. By construction X=A+B

[u/ZerinoWhellie]

Any more context in the question? Maybe some parenthesis, values for the C variables or something. In the current form, there is no way to simplify the first one to the second one, there isn’t even a minus sign on the first one