A normal deck of cards has for each of 13 possible values 2

A normal deck of cards has, for each of 13 possible values (2.. 10, J, K, Q, A), one card with that face value for each of the four suits (hearts, spades, diamonds, clubs). A can either be the highest or lowest value. Compute the number of five card hands (unordered sets of five distinct cards) with the following properties (2 points each, You can leave your answer in terms of factorials or binomial co-efficients. 1 point correct answer, 1 point explanation): The highest value card in the hand is 9 (and aces count higher than 9). Two pairs: there are two pairs with the same value, but no three have the same value. Flush: All cards have the same suit Have at least one card of each suit. Have at least one \"royal\" card (J, K, Q).

Solution

In a deck there are a total of 52 cards = 4*13
now we are supposed to pick 5 cards:
a> cards with higher value than 9 are 10,J,K,Q and A
so a total of = 4*5 = 20 cards have higher value than a 9 in a deck of cards

so total number of cards left to choose 5 cards = 52 - 20 = 32

hence total number of ways = 32*31*30*29*28 = 24165120 ways.

b> we are supposed to make two pairs with the same value and the fifth will be a card of some value other than the two values
whose cards have already been picked

total ways = 52*51*48*47*44 = 263248128 ways

c>
there are 13 cards in a suit and we need a flush
so total ways of getting a flush = 13C5
now there are 4 suits so we\'ll have to times 13C5 be 4 in order to get the final answer
=> total ways = 4*13C5

d>
there will be a suit out of the 4 from which we\'ll have to pick 2 cards as we need a total of 5 cards
number of ways in which 1 card could be pickes from a suit of 13 cards = 13C1
ways if picking 2 cards = 13C2
now any of the four suits could have 2 cards so number of ways of choosing 1 of the 4 suits is = 4C1

hence total ways = 4C1*13C2*(13C1)^3

e>
to have atleast 1 face card
there are a total of = 3*4 = 12 face cards
total number of regular cards = 52-12 = 40
to choose 1 from 12 total ways = 12C1
to choose 2 face cards from 12 total ways = 12C2
to choose 3 face cards from 12 total ways = 12C3
to choose 4 face cards from 12 total ways = 12C4
to choose 5 face cards from 12 total ways = 12C5

total ways of selecting 5 cards = [12C1*40C4 + 12C2*40C3 + 12C3*40C2 + 12C4*40C1 + 12C5]