Permutation ( number of digits)

Question

Question:

(a) How many $5$-digit numbers can be formed from digits $0,1,2,3,4$ if repetition is not allowed?

(b) How many of these are odd?

(c) How many of these are divisible by $5$?

My attempt:

(a) $5!$

(b) $4!(2)=48$

(c) $4! = 24$

Am I correct?

Answer 1

(a) Since you need to get 5 digit numbers using the given digits, there are only 4 possibilities for the first digit - $1,2,3,4$. ( $0$ is not allowed). For the second digit, since repetition is not allowed, there are again $4$ possibilities - $0$ and the remaining $3$ digits. Similarly, $3$ possibilities for the $3$rd digit; $2$ possibilities for the $4$th digit; and $1$ for the $5$th remaining digit; which by multiplication rule, gives $4*4*3*2*1 = 96$; hence there are $96$ possible $5$-digit numbers with the given digits and no repetitions.

(b)In order to get an odd 5-digit number, the last digit is $1$ or $3$. Hence there are $2$ possibilities for the last digit. Then, for the first digit, there are $3$ possibilities, (since again, $0$ is not allowed). For the $2$nd digit, there are $3$ possibilities ( since $0$ is allowed here); $2$ possibilities for the $3$rd digit; and only $1$ possibility for the 4th digit. Hence, there are $3*3*2*1*2 =36$ total odd numbers.

(c) For the $5$-digit number to be divisible by $5$, the last digit is $0$. The first digit has $4$ possibilities ($1,2,3,4$); similarly the $2$nd digit has $3$ possibilities; the $3$rd digit has $2$ possibilities; and the $4$th digit has $1$. Hence there are $4! = 24$ such numbers divisible by 5.