There are 40 people at drake's sweet sixteen party. drake's only rule is that everyone in the party must meet and shake each others hand. if all 40 guest shake each other's hand once and only once. (1) how many handshakes are there altogether? (50 pts) (2) how did you get your solution? no verbiage should be the same for this. there are multiple ways to think about this. (30 pts) (3) what's the quickest way to figure this out if drake invited 1000 people? (20 pts)can someone answer this with an explanation?

1) It's a combination of 2 chosen from 40:
 
⁴⁰C₂ = (40!)/[(2!).(40-2)!] = 780

2) Another way to think about this problem is as follows:

1st shakes hand with 39 (he can't shake his own hand)

2nd shakes hand with 38 (his hand was already shaken by the 1st, so no duplication)

3rd shakes hand with 37 (his hand was already shaken by the 1st, and 2nd, so no duplication)
And so on and so forth...until 1:

Number of hands shaken : 39 + 38 + 37 + 36 + ...+1
The sum = [(1+39).39]/2 = 40x39/2 = 780 
Remember the above is an Arithmetic progression, with first term 1. the last term 39 and the number of terms 39 
S=(a₁+a₃₉)(39/2)

Where’s the problems ? ?