Question

Find the number of numbers of six digits that can be made with the digits 1, 2, 3, 4 if all the digits are to appear in the number at least once

Answer 1

Answer:

Step-by-step explanation:

Out of the six digits,  four digits are fixed.

In the remaining two digits, both the digits could be same  or both the digits could be different as well

Case one - When both the digits are same.

Number of six digit numbers =4C1 × 6! /3!

(Both the digits could be any of the four digits, therefore 4C1)

Case two - When both the digits are different.

(The two digits could be any two of the four digits, therefore 4C2)

Therefore,  the number of six digit numbers in this case = 4C2 × 6!/2!.2!

Thus, the required number of 6 digit numbers = 4C1 × 6!/3! + 4C2 × 6!/2!.2! +

= 6! (23+32)

= 6! × 136

= 1560

Thus, the required number is 1560.