Five cards are dealt off of a standard 52-card deck and lined up in a row. How many such line ups are there in which exactly one of the 5 cards is a queen?
My attempt at a solution:
The queen could be in the first card or the second card or the third card or the fourth card or the fifth card.
There is one queen for each of four suits.
Let $X_{1}$ be the set of 5 cards lined up in a row so that the queen is the first card in that line up.
Let $X_{2}$ be the set of 5 cards line up in a row so that the queen is the second card in that line up.
Let $X_{3}$ be the set of 5 cards line up in a row so that the queen is the third card in that line up.
Let $X_{4}$ be the set of 5 cards line up in a row so that the queen is the fourth card in that line up.
Let $X_{5}$ be the set of 5 cards line up in a row so that the queen is the fifth card in that line up.
Our answer is $|X_{1}|+|X_{2}|+|X_{3}|+|X_{4}|+|X_{5}|$
To find $|X_{1}|$, here is the reasoning:
There are 4 choices for the first card (4 queens to choose from), 51 choices for the second card, 50 choices for the third card, 49 choices for the fourth card and 48 choices for the fifth card. This is so because there are no repetitions of cards because each card is distinct from sny other card. So $|X_{1}| = 4 \times 51 \times 50 \times 49 \times 48$.
A similar process applies to find the cardinality of the four remaining sets.
So
$|X_{2}| = 51 \times 4 \times 50 \times 49 \times 48$
$|X_{3}| = 51 \times 50 \times 4 \times 49 \times 48$
$|X_{4}| = 51 \times 50 \times 49 \times 4 \times 48$
$|X_{5}| = 51 \times 50 \times 49 \times 48 \times 4$.
Therefore
$|X_{1}|+|X_{2}|+|X_{3}|+|X_{4}|+|X_{5}| = 5 \cdot 4 \cdot 51 \cdot 50 \cdot 49 \cdot 48$.
Is my reasoning clear? Are there other different ways of arriving at the correct answer?
Best Answer
Your attempt is not correct.
The question asks for "...line ups are there in which exactly one of the 5 cards is a queen?" but by starting thr multiplication of other cards from $51$, you are including other possible queens, and also overcounting.
If you want to proceed in the manner you have, but using a $5$ multiplier rather than adding $5$ times, the answer would be $5*4*48*47*46*45$
But you must have studied combinations and permutations by now, so a simpler (or more useful) formulation might be
$\binom41\binom{48}4*5!$
[Choose a queen]$\times$[Choose $4$ non-Queens}$\times$ [Permute all $5$]