How many four-digit numbers can be formed with the digits 3, 5, 7, 8, 9 which are greater than 8000, if repetition of digits is not allowed?

Asked by Aaryan | 1 year ago |  26

##### Solution :-

The required numbers are greater than 8000.

So, the thousand’s place can be filled with 2 digits: 8 or 9.

Let us assume four boxes, now in the first box can either be one of the two numbers 8 or 9, so there are two possibilities which is 2C1

In the second box, the numbers can be any of the four digits left, so the possibility is 4C1

In the third box, the numbers can be any of the three digits left, so the possibility is 3C1

In the fourth box, the numbers can be any of the two digits left, so the possibility is 2C1

Hence total number of possible outcomes is 2C1 × 4C1 × 3C1 × 2C1 = 2 × 4 × 3 × 2 = 48.

Answered by Sakshi | 1 year ago

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