Question

How to arrange 2, 2, 2, 1 and 3 in 5 digits.

Answer 1

Hint – In this particular question the total number of ways to arrange n digits number is (n!), later on in the solution use the property if there are any repeated digits in the given digits then the total number of ways to arrange them in the ratio of the (n!) to the factorial the number of times the digit is repeated so use this concept to reach the solution of the question.

Complete step-by-step solution -
Given digits:
2, 2, 2, 1 and 3.
So as we see there are 5 digits available now we have to arrange them in 5 digits.
So first find out the number of ways these digits can be arranged to form a 5 digit number.
So as we know if there are n digits present then the number of ways to arrange them = n!
So there are 5 digits available so the number of ways to arrange them = 5!
Now as we see that in the given digits 2 come three times so the number of ways to arrange them is the ratio of the (5!) to the pictorial times number of times the digit is repeated.
Therefore, the total number of ways these digits can be arranged to form a 5 digit number = $\dfrac{{5!}}{{3!}}$
Now simplify this we have,
$ = \dfrac{{5.4.3.2.1}}{{3.2.1}} = 20$
So there are 20 five digit words possible.
Now the arrangement is
22213, 22231, 22312, 22321, 22132, 22123, 21223, 21232, 21322, 23122, 23212, 23221, 12223, 12232, 12322, 13222, 31222, 32221, 32212, 32122.
So these are the required arrangements.

Note – Whenever we face such types of questions the key concept we have to remember if there are n digits in the system in which r digits are same then the number of ways to arrange them is $\dfrac{{n!}}{{r!}}$, if there are two types of digits repeated (i.e. one type of r digits are same and another type of p digits are same), then the number of ways to arrange them is $\dfrac{{n!}}{{r!.p!}}$.