A four-digit number is formed by using only the digits 1, 2 and 3 such that both 2 and 3 appear at least once. The number of all such four-digit numbers is

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Text Explanation

The question asks for the number of 4 digit numbers using only the digits 1, 2, and 3 such that the digits 2 and 3 appear at least once.

The different possibilities include :

Case 1:The four digits are ( 2, 2, 2, 3). Since the number 2 is repeated 3 times. The total number of arrangements are :

$\frac{4!}{3!}$ = $4$.

Case 2: The four digits are 2, 2, 3, 3. The total number of four-digit numbers formed using this are :

$\frac{4!}{2!(2!)}$ = $6$.

Case 3: The four digits are 2, 3, 3, 3. The number of possible 4 digit numbers are :

$\frac{4!}{3!}$ = $4$.

Case4: The four digits are 2, 3, 3, 1. The number of possible 4 digit numbers are :

$\frac{4!}{2!}$ = $12$.

Case5: Using the digits 2, 2, 3, 1. The number of possible 4 digit numbers are :

$\frac{4!}{2!}$ = $12$.

Case 6: Using the digits 2, 3, 1, 1. The number of possible 4 digit numbers are :

$\frac{4!}{2!}$ = $12$.

A total of 12 + 12 + 12 + 4 + 6 + 4 = 50 possibilities.

Alternatively,

We have to form 4 digit numbers using 1,2,3 such that 2,3 appears at least once

So the possible cases :

Now we get $\frac{4!}{2!}$ × 3 = 36 ( When one digit is used twice and the remaining two once )

$\frac{4!}{3!}$ × 2 = 8 ( When 1 is used 0 times and 2 and 3 is used 3 times or 1 time )

$\frac{4!}{2! × 2!}$ = 6( When 2 and 3 is used 2 times each )

So total numbers = 36+8+6 = 50

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