Very difficult Math question from Chung's book

purpleacorn
Fri, 19 Apr 2013 10:27:44 +0000

There are seven slots, but there are limitations as to what can go in each slot.

The first card can either be an A or a G, so there are two choices. Similarily, the last card can only be what is left over from the pair, so there is only one choice.

After that, we have a couple of possibilities on the middle five cards.

Starting with the second slot, there are five options for what that card can be (since BCD does not have to stay in that order.

IF the second card is a B, C, or D, then the third card must be one of that triplet, and the fourth card must be one as well. (We will deal with the other option later.)

Then the fifth card can either be E or F, and the sixth card will be the last.

That’s a multiplication of 2321211, or 24 arrangements.

Now, what if the second card isn’t a B, C, or D? But the third is one of those, and then the fourth and fifth are similarily a triplet card? The sixth card would still be a single card.

So 2232111, or another 24.

And then you need to consider if the B, C, D triplet doesn’t come until the fourth, fifth, and sixth cards (2213211).

I’m getting 24*3, which is 72, mainly because we do not have a limit that BCD stay together *in that order/i, etc.