AMC 10 Weekly Practice Round 3

Complete problem set with solutions and individual problem pages

Problem 6 Medium

The number of six-digit numbers formed from the digits 0,1,2,3,4,5 in which odd and even digits alternate and no digit is repeated is （   ）．

  • A.

    72

  • B.

    60

  • C.

    48

  • D.

    36

  • E.

    12

Answer:B

The total number of alternating arrangements of odd and even digits using these six digits is 2 \left( _{3}P_{3} \right)^{2} = 72.

 

Among them, the arrangements with 0 in the first place are _{3}P_{3} \times _{2}P_{2} = 12.

 

Therefore, the number of six-digit numbers satisfying the condition is 72 - 12 = 60.

 

Thus, the correct answer is \text{B}.

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