Six cards are to be selected from a deck of cards. At least 1 card is to be selected from each suit. In how many different ways can this be done?
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4C1 × 13C3 × (13C1)3
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(4C2) × (13C2)2 × (13C1)2
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4C1 × 13C3 × (13C1)3 + 4C2 × (13C2)2 × (13C1)2
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4C1 × 13C3 × (13C1)3 + (4C2 ÷ 2) × (13C2)2 × (13C1)2
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None of the above
To select 6 cards with at least one from each suit, the distribution of suits can be (3,1,1,1) or (2,2,1,1). For (3,1,1,1), choose 1 suit for 3 cards (4C1), choose 3 cards from it (13C3), choose 1 card from each of the other 3 suits (13C1^3). For (2,2,1,1), choose 2 suits for 2 cards (4C2), choose 2 cards from each (13C2^2), choose 1 card from each of the other 2 suits (13C1^2).
The selection can be split into two valid distribution cases: either one suit provides three cards and the others provide one, or two suits provide two cards and the others provide one. For the first case, choose the suit for three cards and select them, giving 4C1 * 13C3 * (13C1)^3. For the second case, choose the two suits for two cards and select them, giving 4C2 * (13C2)^2 * (13C1)^2. Adding these gives the total expression 4C1 * 13C3 * (13C1)^3 + 4C2 * (13C2)^2 * (13C1)^2.