SOLUTION: In how many ways can a person get a bridge hand consisting of two aces;one king;one queen;three jacks;and the six other cards ten or less?Please help me to solve this question.

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Question 936253: In how many ways can a person get a bridge hand consisting of two aces;one king;one queen;three jacks;and the six other cards ten or less?Please help me to solve this question.
Found 2 solutions by Theo, AnlytcPhil:
Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
in one deck of cards there are:

4 aces (1 for each suit)
4 kings (1 for each suit)
4 queens (1 for each suit)
4 jacks (1 for each suit)
36 cards from 2 to 10 (9 for each suit)

the suits are spades, clubs, hearts, diamonds.

the number of ways to get each one of these is:

4c2 for 2 aces from 4 aces = 6
4c1 for 1 king from 4 kings = 4
4c1 for 1 queen from 4 queens = 4
4c3 for 3 jacks from 4 jacks = 4
36c6 for 6 cards from 36 cards that are numbered from 2 to 10 = 1,947,792

the number of possible ways you can get 2 aces, 1 king, 1 queen, 3 jacks, 6 other cards is:

6 * 4 * 4 * 4 * 1,947,792 = 759,472,128

ncx equals the combination formula to get sets of x from n which is equal to n! / (x! * (n-x)!)

for example:

4c2 = 4! / 2! * 2! = (4*3*2*1) / (2*1*2*1) = 24 / 4 = 6

the assumption with the cards from 2 to 10 is that any combination of 6 of them can be part of your hand without any restrictions, i.e., you could have 4 sevens and 2 eights or you can have 2 sevens and 2 eights and 2 nines, oryou can have a straight flush, etc.


Answer by AnlytcPhil(1806)   (Show Source): You can put this solution on YOUR website!

two aces;
4 aces Choose 2

That's 4C2 = 6 ways

one king;
4 kings Choose 1

That's 4C1 = 4 ways

one queen;
4 queens Choose 1

That's C(4,1) = 4 ways

three jacks;
4 jacks choose 3

That's C(4,3) = 4 ways

and the six other cards, ten or less
 There are 9 ranks (2,3,4,5,6,7,8,9,10) of 4 suits each. 
 That makes 9*4=36 cards which are ten or less.

36 cards ten or less Choose 6

That's C(36,6) = 1947792

Answer: (4C2)(4C1)(4C1)(4C3)(36C6) = (6)(4)(4)(4)(1947792) = 747952128 ways.

Edwin
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