Question

1. consider a standard deck of 52 cards. determine the number of distinct six card hands that are possible which include

a) no restrictions

b) only clubs

c) 2 clubs and 4 diamonds

d) no sevens

e) 4 tens

f) exactly 1 jack and 4 queens

Explanation:

Problem 6 (Face Cards Removed, Three - Card Hands)

Part a: No Restrictions

Step 1: Determine the number of non - face cards

A standard deck has 52 cards. There are 12 face cards (4 Jacks, 4 Queens, 4 Kings). So the number of non - face cards is \(52 - 12=40\).

Step 2: Calculate the number of 3 - card hands

We use the combination formula \(C(n,r)=\frac{n!}{r!(n - r)!}\), where \(n = 40\) (number of non - face cards) and \(r = 3\) (number of cards in a hand).
\(C(40,3)=\frac{40!}{3!(40 - 3)!}=\frac{40!}{3!×37!}=\frac{40\times39\times38}{3\times2\times1}=9880\)

Step 1: Determine the number of Kings

There are 4 Kings in a standard deck. But we want to choose 3 Kings.

Step 2: Calculate the number of combinations

Using the combination formula \(C(n,r)\) with \(n = 4\) (number of Kings) and \(r = 3\) (number of Kings in the hand).
\(C(4,3)=\frac{4!}{3!(4 - 3)!}=\frac{4!}{3!×1!}=\frac{4\times3!}{3!×1}=4\)

Step 1: Choose 1 Queen

There are 4 Queens. The number of ways to choose 1 Queen is \(C(4,1)=\frac{4!}{1!(4 - 1)!}=\frac{4!}{1!×3!}=4\)

Step 2: Choose 2 Kings

There are 4 Kings. The number of ways to choose 2 Kings is \(C(4,2)=\frac{4!}{2!(4 - 2)!}=\frac{4\times3\times2!}{2!×2!}=\frac{4\times3}{2\times1}=6\)

Step 3: Use the multiplication principle

The total number of hands with 1 Queen and 2 Kings is the product of the number of ways to choose the Queen and the number of ways to choose the Kings. So \(C(4,1)\times C(4,2)=4\times6 = 24\)

Answer:

\(9880\)

Part b: 3 Kings