Difficult quant brainteaser need help
You're at the shooting range and have a 3x3 array of targets. You can only shoot the bottom targets of each column first before you can shoot any higher targets in that column. So let's say you shoot target at position (3,3), then you are allowed to shoot the target at position (3,2) but not targets at positions (3,1) or (2,2). This is because after shooting the target at (3,3), the target at (3,2) is now the bottommost target in that column but there is still a target at (2,3). How many different ways can you shoot down the targets?
It’s a nice teaser.
You have 9 targets. If you didn’t have any restriction, you could shoot them in 9! different ways. That’s 362,880 ways.
But you are given the restriction that each column can only be shot sequentially. So basically each column (which can be shot in 3! ways, that’s 6 ways). So you need to divide by that.
The answer is 9! / (3!x3!x3!) = 362,880 / 216 = 1,680 ways.
You can do the same math on a 2x2 array: 4! / (2!x2!) = 6, which will help you corroborate the answer in a more tractable way.
Or in a 2x3 array (2 rows, 3 columns) 6! / (2!x2!x2!) = 90 ways
Or a 3x2 6! / (3!x3!) = 20 ways
Very dumb question... how would you calculate 9! in an interview?
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