If the first and last digit must be a 2 or a 4, how many 5-digit numbers are possible?
oa: 4,000
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combinatorics 2
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surajgarg
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The middle 3 digits can be any digit from 0 to 9.
That means - 10x10x10 = 1000
The first and last digits can have combination as
2 _ _ _ 2
2 _ _ _ 4
4 _ _ _ 4
4 _ _ _ 2
4 combinations.
Total 5 digit combinations possible = 1000 x 4 = 4000
That means - 10x10x10 = 1000
The first and last digits can have combination as
2 _ _ _ 2
2 _ _ _ 4
4 _ _ _ 4
4 _ _ _ 2
4 combinations.
Total 5 digit combinations possible = 1000 x 4 = 4000
