Combination (stats) question

1. How many different hands of 13 cards be dealt among 4 players from a deck of 52 cards?
2. How many sums can be formed from the numbers 1, 2, 3, 4, 5?

-- explanations please!

1) selecting 13 cards for first hand can be done in 52C13
for second hand = 39C13
for third hand = 26C13
it can be distributed to four hands in 4! ways
required ways = (52C13)*(39C13)*(26C13)*4!

2) total numbers formed = 5! = 120
required sum = 24*(15)*(10^4 + 10^3 + 10^2 + 10 + 1) = 3,999,960

1.) It doesn't matter how the cards are dealt, so we can just look at the case where
the first player gets dealt 13 of 52 cards in (52 C 13) ways, the second player gets
dealt 13 of the remaining 39 cards in (39 C 13) ways, the third player gets dealt
13 of the remaining 26 cards in (26 C 13) ways, and the fourth player gets the
remaining 13 cards. So the total number of hands is

(52 C 13)*(39 C 13)*(26 C 13)*(13 C 13) = 52! / (13!)^4 = 5.36447 * 10^28

to 6 significant figures.

2.) The numbers can be summed to any value from 1 to 15.

Edit: If order does not matter and if a sum must involve at least two terms,
then there are (5 C 2) + (5 C 3) + (5 C 4) + (5 C 5) = 26 distinct sums.

Note that another way of looking at #1) is to line up the 52 cards and then
assign 13 each of letters A, B, C and D to the cards, where the letters
represent the 4 players. Then this becomes a permutations with
repetition of indistinguishable objects problem, and thus has the
solution 52! / (13!)^4 as before. This also shows that it doesn't actually
matter how the cards are dealt, just that each player eventually gets
13 cards.

1. Hand A can have 52C13 possibilities leaving 39C13 for hand B, 26C13
for hand C and 1 for hand D so in total (52C13)(39C13)(26C13)
=5.364473777x10^28, rather a lot.

2. I assume the numbers cannot be chosen more than once.
If you mean the number of ways one or more of the numbers can be
added it is 5C1+5C2+5C3+5C4+5C5=31
If you mean how many different totals you need a different strategy.

1.
there are 4 players so you times 4 x (52C13)
2
there are 5 numbers so
5C2 + 5C3 +5C4 + 5C5