Question

How many 6-digit numbers are there such that 

(a) the digits of each number are all form the set {1,2,3,4,5} 

(b) any digit that appears in the number appears at least twice.

(Example : 225252 is an admissible number ,while 222133 is not.)

Answer 1

Explanation:

Since each digit occurs at least twice, we have following possibilities : 

1. Three digits occurs twice each. We may choose three digits from {1, 2, 3, 4, 5} in (⁵₃C) = 10ways. If each occurs exactly twice, the number of such admissible 6-digit numbers is

6/2!2!2! x 10 = 900.

2. two digits occur three times each. We can choose 2 digits in (⁵₂C)  = 10ways. Hence the number of admissible 6-digit numbers is

6!/3!3! x 10 = 200.

3.One digit occurs four times and the other twice. We choosing two digits again which can be done in 10 ways. The two digits are interchangeable. Hence the desired number of admissible 6-digit numbers is

2 x 6!/4!2! x 10 = 300.

4. Finally, all digits are same. There are 5 such numbers. Thus the total number of admissible numbers is 900 + 200 + 300 + 5 =1405