Question -
Answer -
The required numbersare greater than 8000.
So, the thousand’splace can be filled with 2 digits: 8 or 9.
Let us assume fourboxes, now in the first box can either be one of the two numbers 8 or 9, sothere are two possibilities which is 2C1
In the second box, thenumbers can be any of the four digits left, so the possibility is 4C1
In the third box, thenumbers can be any of the three digits left, so the possibility is 3C1
In the fourth box, thenumbers can be any of the two digits left, so the possibility is 2C1
Hence total number ofpossible outcomes is 2C1 × 4C1 × 3C1 × 2C1 =2 × 4 × 3 × 2 = 48.