How many different nine digit numbers can be formed from the number 66 55 99 222 by rearranging its digits so that the odd digits occupy even positions only?
Number of even places = 4 Number of even digits = 5 (6,6,2,2,2) Number of odd places = 5 Number of odd digits = 4 (5,5,9,9) Since 5 & 9 are repeated two times odd digits can be arranged in 4!/(2!×2! ) = 6 ways Since 6 is repeated two times & 2 is repeated three times even digits can be arranged in 5!/(2! ×3!) = 10 ways Hence, the required number of ways = 6 × 10 = 60 ways.
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