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Permutations and Combinations!

how many 5 digit numbers can be formed that do not contain the digits 0 or 8?

User Aushin
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1 Answer

3 votes

Answer:

9×9×8×7×6= 27,216


Explanation:

Your answer is :- 9×9×8×7×6= 27,216

Why did i multiply 9×9×8×7×6 ⤵️

Because we can fill each place of a five digit number with ten digits (0,1,2,3,4,5,6,7,8,9).

Now, let us assume the five digit number is abcde, where each alphabet represent a digit.

so according to our question a cannot be equal to 0 ( a≠0), because if a will be equal to zero then our number will be a four digit number. So a can be equal to 9 digits.

b digit can also be equal to 9 numbers. why??because we can place zero on the position of b and only one number will not be repeated which is a.

Now for c, d, e :- accordingly no number will be repeated so on the place of c, there can be 8 digits. On the place of d, there can be 7 digits. And on the place of e there can be 6 digits.

Therefore,

a×b×c×d×e=9×9×8×7×6=27,216


User Nola
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7.3k points