The number of 9 digit numbers formed using the digit 223355888 such that odd digits occupy even places is (a) 16 (b) 36 (c) 60 (d) 80

We have three odd digits (two 3s and one 5) that must be placed in the even positions (2nd, 4th, 6th, and 8th). We need to find all possible ways to arrange these odd digits in these even positions. We can do this by using the combination formula: \(C(n,r) = \frac{n!}{(n-r)!r!}\) where n is the total number of items, r is the number of items you want to choose, and ! denotes a factorial (e.g., 5! = 5 × 4 × 3 × 2 × 1). There are 4 even positions, and we need to choose 3 of these positions for the odd digits. So, n = 4 and r = 3. Let's calculate the combinations: \(C(4,3) = \frac{4!}{(4-3)!3!} = \frac{4!}{1!3!} = \frac{4 × 3 × 2 × 1}{(1)(3 × 2 × 1)} = 4\) There are 4 different ways to place the odd digits (two 3s and one 5) in the even positions.

Since we already placed the odd digits in the even positions, we are now left with 5 odd positions (1st, 3rd, 5th, 7th, and 9th) for the even digits (two 2s and three 8s). We can simply calculate the number of ways to arrange the even digits in these odd positions using a similar approach. There are 5 odd positions, and we need to choose 2 of these positions for the two 2s. So, n = 5 and r = 2. Let's calculate the combinations: \(C(5,2) = \frac{5!}{(5-2)!2!} = \frac{5!}{3!2!} = \frac{5 × 4 × 3 × 2 × 1}{(3 × 2 × 1)(2 × 1)} = 10\) There are 10 different ways to place the even digits (two 2s and three 8s) in the odd positions.

To find the total number of possible arrangements, we multiply the number of ways to place the odd digits (4) by the number of ways to place the even digits (10): Total arrangements = 4 × 10 = 40 However, in our current count, we have considered the numbers where the two 3s or the two 2s are together as different numbers. We need to adjust our count by eliminating the cases where the two 3s or the two 2s are together. In this case, there are 2 ways to arrange the 3s in 2nd and 4th position and 2 ways to arrange the 3s in 6th and 8th position, giving a total of 4 additional ways. Final count of arrangements = Total arrangements + Additional ways = 40 + 4 = 44 However, 44 is not one of the options presented in the exercise. Reviewing our calculations, we realize that our adjustment was incorrect. Since odd digits should occupy even places, having two identical odd digits together is impossible. So, there is no need to adjust our initial count. Final count of arrangements = Total arrangements = 40 The number of 9-digit numbers formed using the digit 223355888 such that odd digits occupy even places is not among the options given. The closest answer is either: (e) None of the above, or (f) 40 Both options indicate that the correct answer is not among the given choices.