How many distinct 8-digit numbers can be formed by rearranging the digits of the number 11223344 such that odd digits occupy odd positions and even digits occupy even positions?
Answer & explanation
Answer: (c) 36
The four odd positions must hold 1, 1, 3, 3 and the four even positions must hold 2, 2, 4, 4. Each set can be arranged in 4!/(2! × 2!) = 6 ways, so there are 6 × 6 = 36 numbers.
- Odd digits 1, 1, 3, 3 fill the 4 odd positions; even digits 2, 2, 4, 4 fill the 4 even positions.
- Arrangements of 1, 1, 3, 3 = 4!/(2! × 2!) = 24/4 = 6.
- Arrangements of 2, 2, 4, 4 = 6 in the same way.
- Total = 6 × 6 = 36.
Remember · Arrangements with repeated items = n!/(p! × q! …); choices for independent slots multiply.
Question and answer: UPSC's official GS Paper II (2023, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). ·