Question:
8-digit numbers are formed using the digits \(1, 1, 2, 2, 2, 3, 4, 4\). The number of such numbers in which the odd digits do not occupy odd places, is:
8-digit numbers are formed using the digits \(1, 1, 2, 2, 2, 3, 4, 4\). The number of such numbers in which the odd digits do not occupy odd places, is:
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The odd digits (1, 1, 3) must go into the 4 even places; arrange them there and then arrange the even digits in the remaining places.
Updated On: Nov 11, 2026
- 160
- 120
- 60
- 48
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The Correct Option is B
Solution and Explanation
The odd digits are \(1, 1, 3\) (three of them) and the even digits are \(2, 2, 2, 4, 4\) (five of them).
There are 4 odd places (1st, 3rd, 5th, 7th) and 4 even places (2nd, 4th, 6th, 8th). The odd digits must not be in odd places, so all three go into even places.
Ways to place \(1, 1, 3\) in the 4 even places: \(\frac{4!}{2!\,1!\,1!} = 12\) (3 can be put in 4 ways, then the two 1's in \(\binom{3}{2} = 3\) ways; \(4 \times 3 = 12\)).
The remaining 5 places (4 odd places and 1 even place) are filled with \(2, 2, 2, 4, 4\): \(\frac{5!}{3!\,2!} = 10\) ways.
Total \(= 12 \times 10 = 120\).
Answer: (2) 120
There are 4 odd places (1st, 3rd, 5th, 7th) and 4 even places (2nd, 4th, 6th, 8th). The odd digits must not be in odd places, so all three go into even places.
Ways to place \(1, 1, 3\) in the 4 even places: \(\frac{4!}{2!\,1!\,1!} = 12\) (3 can be put in 4 ways, then the two 1's in \(\binom{3}{2} = 3\) ways; \(4 \times 3 = 12\)).
The remaining 5 places (4 odd places and 1 even place) are filled with \(2, 2, 2, 4, 4\): \(\frac{5!}{3!\,2!} = 10\) ways.
Total \(= 12 \times 10 = 120\).
Answer: (2) 120
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