CAT 2021

Slot 3

QA

Question & Solution

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

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Case 1: Digits = (2, 2, 2, 3)
Number 2 repeats three times, so the arrangements are: $$\frac{4!}{3!} = 4$$

Case 2: Digits = (2, 2, 3, 3)
Arrangements: $$\frac{4!}{2!,2!} = 6$$

Case 3: Digits = (2, 3, 3, 3)
Arrangements: $$\frac{4!}{3!} = 4$$

Case 4: Digits = (2, 3, 3, 1)
Arrangements: $$\frac{4!}{2!} = 12$$

Case 5: Digits = (2, 2, 3, 1)
Arrangements: $$\frac{4!}{2!} = 12$$

Case 6: Digits = (2, 3, 1, 1)
Arrangements: $$\frac{4!}{2!} = 12$$

Total: $12 + 12 + 12 + 4 + 6 + 4 = \boxed{50}$

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Alternate Method:

We form 4-digit numbers using digits 1, 2, 3 such that 2 and 3 appear at least once.

Possible patterns:

1. One digit used twice and the remaining two digits once each
   Ways: $$\frac{4!}{2!} \times 3 = 36$$
   (three choices for the digit that appears twice)

2. One digit used 3 times, the other used once
   Only possible when digits 2 and 3 appear:
   Patterns → (2,2,2,3) and (2,3,3,3)
   Ways: $$\frac{4!}{3!} \times 2 = 8$$

3. Digits 2 and 3 appear twice each
   Ways: $$\frac{4!}{2!\times2!} = 6$$

Total: $$36 + 8 + 6 = 50$$

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Final Answer: $\boxed{50}$