How many four-digit numbers can be formed with the digits 3, 5, 7, 8,

1.	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?
Answer» Given : Four-digit number is required which is greater than 8000 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 Similarly, For the third box, the numbers can be any of the three digits left, so the possibility is ³C₁ For 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

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?

Answer»

Given :

Four-digit number is required which is greater than 8000

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

Similarly,

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

For 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