Statistics Problem Permutations and Combinations

Question

Six cards are dealt from a standard deck of 52, and placed in a sequence. this set, and the subsets listed below, can be counted using trees. You should give your answer as a product, to indicate how you found the formula. A. How many sequences are there? (Note that no card appears twice in any sequence.) B. How many sequences are there in which no card is a Spade? C. How many sequences are there in which no two cards have the same face value? D. How many sequences are there in which exactly one card is a Jack? I really need an explanation to this. This is What I currently have A. 52P6. B. 39P6. C. 13*12*11*10*9*8. I don't know for the last one.

Answer 1

Alternatively to Guacho's explanation for part $D$ approach via multiplication principle:

• Pick which of the six spots in the sequence has a jack: $6$ options
• Pick which specific jack it is: $4$ options
• From left to right, pick a non-jack card to fill each remaining position in the sequence: $48\cdot 47\cdot 46\cdot 45\cdot 44$ options, i.e. $~_{48}P_5$ options

Multiplying, we got a total of $6\cdot 4\cdot ~_{48}P_5$ sequences of six cards with exactly one jack.

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As a further aside, your answer to part $C$ is incorrect. Your answer would be correct if it was "no two have the same face value and they are all spades." You must account for what suits are possible for each card in the sequence, increasing the total count by an additional $4^6$.

Answer 2

For the last one, there are $C(4,1)$ ways to choose the card that will be a jack, and there are $C(48,5)$ to choose the remaining $5$ cards from the $48$ cards in the deck that are not jacks. Then the answer is $$6!C(4,1)C(48,5)$$ Where the $6!$ comes because for every choice of $6$ cards, there are $6!$ ways of arranging said choice.