Question

how many 4 digit numbers are there which have exactly 3 distinct digits?​

Answer 1

Step-by-step explanation:

There are 9×9×8×7 4-digit numbers that have 4 different digits. There are 9×9×8 4-digit number that have 3 different digits. But this 3 different digits can be place on 6 different ways: aabc,abac,abca,baac,baca,bcaa, so we multiply the number of combintion by 6.

Answer 2

Permutation and Combination

We have to take digits from $0,1,2,3,4,5,6,7,8,9$

Hence we have total of 10 digits. The condition here is exactly 3 digits should be distinct and one more condition automatically applied here is that 1st digit can't have 0.This means 2 digits will be repeater and 2 will be distinct.

The three case will be formed :

1) when  unit digit is repeated

The way of select 1st three digits are $9\times9\times8 \ ways$.

(∵ 1st place have only 9 option except 0, 2nd place has 9 options except the digit at 1st place, 3td place have 8 option excluding 1st place and 2nd place.)

The 4th place can be same as either of the three digits used earlier , hence it will have 3 ways.

Total number of ways becomes $=9\times 9\times 8\times 3$ ways.

2) when  ten's digit is repeated

first, third and fourth digit can be selected in $9\times 9\times 8\ ways$.

The 3rd place can be same as first 2 digits, earlier, hence it will be have 2 ways.

Total number of ways becomes $=9\times9\times2\times8$ ways.

3) when  hundred's digit is repeated

first, third and fourth digit can be selected in $9\times 9\times 8\ ways$.

The 2nd place can be same as first digit, earlier, hence it will be have 1 ways.

Total number of ways becomes $=9\times1\times9\times8$ ways.

Total number of ways will be

$9\times9\times8\times 3+9\times9\times2\times 8+9\times1\times9\times 8\\\\=(9\times9\times8)(3+2+1)\\\\=9\times9\times8\times6=3888 \ ways$

Hence required number of ways is 3888.