# Thread: Counting Principle 52-card deck

1. ## Counting Principle 52-card deck

How many ways are there to pick 2 different cards from a standard 52-card deck such that:

The fist card is a spade and the second card is not a Queen

The answer is (1*48) + (12*47)

I do not know how to get there again.

I have 13 spade cards in a 52-card deck
Ace, 2,3,4,5,6,7,8,9,10, Jack, Queen, King

so I thought there are 13 possibilities for the first card
For the second card we cannt have Queens (4) and one card is already gone,
so that would be 52-4 - 1 = 47

I see the problem that maybe for the first card a spade Queen was used, but I do not know how to set the problem up with this information.

Thanks for any hint.

2. Hello, DBA!

How many ways are there to pick 2 different cards
from a standard 52-card deck such that:

The first card is a Spade and the second card is not a Queen.

The answer is: .$\displaystyle (1\cdot48) + (12\cdot47)$

We must consider two cases:

. . (1) the first card is the $\displaystyle Q\spadesuit.$

. . (2) the first card is not the $\displaystyle Q\spadesuit.$

(1) The first card is the $\displaystyle Q\spadesuit$.
. . . There is one way to accomplish that.

. . . Among the other 51 cards, there are 3 Queens and 48 Others.
. . . And we want one of the Others: 48 ways.

. . . Hence: .$\displaystyle n(\text{1st}\,Q\spadesuit \:\wedge \sim\!Q) \;=\;1\cdot48$

(2) The first card is a $\displaystyle \spadesuit$, but not the $\displaystyle Q\spadesuit.$
. . . There are 12 ways to accomplish that.

. . . Among the other 51 cards, there are 4 Queens and 47 Others.
. . . And we want one of the Others: 47 ways.

. . . Hence: .$\displaystyle n(\text{1st} \sim\! Q\spadesuit\;\wedge\,\sim\!Q) \;=\;12\cdot47$

Therefore: .$\displaystyle n(\text{1st}\spadesuit\;\wedge\:\sim\!Q) \;=\;1\cdot48 + 12\cdot47 \;=\;612\text{ ways.}$

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### how many ways are there to pick 2different cards from a standard 52 card deck such that the first card is a spade and the second card is not a queen

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