Number of combinations

You must strictly need all numbers in ${2, 4, 6, 8}$ . Therefore minimum of 4 digit in 5 digit number is filled.

$\text{First 4 digit is filled, and last digit is chosen at random, so the number would be in the form} \\ \boxed{\text{a}}\boxed{\text{b}}\boxed{\text{c}}\boxed{\text{d}}(\boxed{a/b/c/d}) \\ \text{a, b, c, d can choose all 4 numbers in {2, 4, 6, 8} and d can choose any 1 number from {2, 4, 6, 8}} \\ \therefore \text{Total no of ways to form 5 digit number = } 5! \\ \text{Now the digits can be repeated, in first 4 digits and last digits together. So, when digits repeat 2 times we have} \\ \text{No of ways = } \dfrac{5!}{2!} \\ \text{Now, the last digit can be filled in 4 ways} \\ \boxed{\implies \text{Total no of ways to form such 5 digit number } = \dfrac{5!}{2!} \times 4 = 240}$

we have to put all the digits for this number. Then for the set = { 2,4,6,8} we need to have all the digits and one digit would be extra in all. So the numbers will have all the 4 digits from {2,4,6,8} and another from the same set but only one like 22648 ... now this type of number with two two's can be permutated in 5!/2! ways... isnt it? Then for two fours we have 5!/2! and for the all four we have the same. Hence Total ways = 5!/2! x 4 = 240 .... And C= 240.