Breaking Down the Info:

We have 9 digits in total, so there are 5 odd places. Among the digits, 3, 5, 3, 5, and 3 are odd so we should slot those first. We need to slot in the 3's (or 5's) first and the rest of the digits are confirmed, so it is 5C3 positions for the odd places.

Then we have to deal with the even places as well. The even digits are 6, 6, 2, and 4. We can randomize for 4! placements and divide by 2! to eliminate duplicates due to the 6's.

Putting everything together, it is \(5C3 * \frac{4!}{2!} = \frac{5*4}{2} * 4 * 3 = 120\) options.

Answer: B
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Could anyone kindly solve this please?

ashishpathak

We have 3,3,3,5,5 to put into the five odd-numbered slots and 2,4,6,6 to put into the four even-numbered slots.

Odds:
If we know what we're doing, we could use \(\frac{5!}{3!2!} = \frac{5*4}{2} = 10\).
But even if we don't, it's probably just as fast to simply count where the 5s could go (the 3s go in the other slots). The 5s could go in slots 1&3, 1&5, 1&7, 1&9, 3&5, 3&7, 3&9, 5&7, 5&9, 7&9. That's 10 ways for the odds.

Evens:
If we know what we're doing, we could use \(\frac{4!}{2!} = 4*3 = 12\).
But even if we don't, we can reason that there are 4*3*2*1 ways to place four things into the four even slots, and then we need to divide by 2 since the 6s are interchangeable, so 4*3 = 12 ways for the evens.

Multiply those together and we have 10*12 = 120.

Answer choice B.
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