How many four-digit numbers are possible in which the leftmost digit is odd, the rightmost digit is even, and all four digits are different

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asked Jun 18, 2017 128k views

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Srohde · Answer 1 · 2017-06-21T20:47:00+0000

Answer:

1400

Explanation:

Count from left to right: There are 5 choices for the first digit, 5 choices for the second, 8 remaining choices for the third, and 7 remaining for the fourth, so there are $5*5*8*7= 1400

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How many four-digit numbers are possible in which the leftmost digit is odd, the rightmost digit is even, and all four digits are different

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