Counting ll 2. As an extra check, try solving the same problem in a different way. Multiple ap- proaches are often available-and all had better give the same answer! (Sometimes different approaches give answers that look different, but turn out to be the same after some algebra. We already used the first method; let's try the second. There is a bijection between hands with two pairs and sequences that specify 1. The values of the two pairs, which can be chosen in(2)ways 2. The suits of the lower-value pair, which can be selected in(2)ways 3. The suits of the higher-value pair, which can be selected in(2)ways 4. The value of the extra card, which can be chosen in 11 ways 5. The suit of the extra card, which can be selected in(0)=4 ways For example, the following sequences and hands correspond: ({3,Q},{◇,命},{◇,9},A,品 3◇,3·,Q◇,Q,A ({9,5},{,},{,◇},K,◆)→{99,9,59,5晶,K命 Thus the number of hands with two pairs is: 11·4 This is the same answer we got before, though in a slightly different form 3.4 Hands with Every Suit How many hands contain at least one card from every suit? Here is an example of such a hand {7◇,K昴,3◇,A9,2·} Each such hand is described by a sequence that specifies 1. The values of the diamond the club, the heart and the spade which can be selected in13.13·13·13=134ways 2. The suit of the extra card, which can be selected in 4 ways 3. The value of the extra card, which can be selected in 12 ways� � � � � � � � 8 Counting III 2. As an extra check, try solving the same problem in a different way. Multiple ap￾proaches are often available— and all had better give the same answer! (Sometimes different approaches give answers that look different, but turn out to be the same after some algebra.) We already used the first method; let’s try the second. There is a bijection between hands with two pairs and sequences that specify: 13 1. The values of the two pairs, which can be chosen in ways. 2 2. The suits of the lower­value pair, which can be selected in 4 2 ways. 3. The suits of the higher­value pair, which can be selected in 4 2 ways. 4. The value of the extra card, which can be chosen in 11 ways. 5. The suit of the extra card, which can be selected in 4 1 = 4 ways. For example, the following sequences and hands correspond: ({3, Q} , {♦, ♠} , {♦, ♥} , A, ♣) ↔ ({9, 5} , {♥, ♣} , {♥, ♦} , K, ♠) ↔ { { 3♦, 9♥, 3♠, 9♦, Q♦, 5♥, Q♥, 5♣, A♣ K♠ } } Thus, the number of hands with two pairs is: � � � � � � 13 4 4 2 · 2 · 2 · 11 · 4 This is the same answer we got before, though in a slightly different form. 3.4 Hands with Every Suit How many hands contain at least one card from every suit? Here is an example of such a hand: { 7♦, K♣, 3♦, A♥, 2♠ } Each such hand is described by a sequence that specifies: 1. The values of the diamond, the club, the heart, and the spade, which can be selected in 13 · 13 · 13 · 13 = 134 ways. 2. The suit of the extra card, which can be selected in 4 ways. 3. The value of the extra card, which can be selected in 12 ways