Arrangement with repetitions

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 :

$$\frac{4!}{2!\cdot2!}=\ 6$$

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 $$\frac{4!}{2!}\times\ 3$$= 36 ( When one digit is used twice and the remaining two once )
$$\frac{4!}{3!}\times\ 2$$ = 8 ( When 1 is used 0 times and 2 and 3 is used 3 times or 1 time )
$$\frac{4!}{2!\times\ 2!}=\ 6$$( When 2 and 3 is used 2 times each )
So total numbers = 36+8+6 =50

The number of 4-digit numbers possible using 1,1,2, and 4 is $$\frac{4!}{2!}=12$$

Number of 1's, 2's and 4's in units digits will be in the ratio 2:1:1, i.e. 6 1's, 3 2's and 3 4's.

Sum = 6(1) + 3(2) + 3(4) = 24

Similarly, in tens digit, hundreds digit and thousands digit as well.

Therefore, sum = 24 + 24(10) + 24(100) + 24(1000) = 24(1111)

Mean = $$\frac{24\left(1111\right)}{12}=2222$$

The answer is option A.

The formula for arrangement with repetition is n!/(a!b!c!) where n items contain a, b and c repetitions. In this case the letter P is repeated twice and E thrice. Hence the formula is 11!/(2!3!) = (11!/3!)/2 = $$ ^{11}\textrm{P}_{8}/2$$