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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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lchukwu001
1188 is my answer malee2021 sheesh ☺

Answer:

1400

Step-by-step 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

:I

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