## 11/23/09

### Test 4 Section 6 - #18 (page 598)

There are two ways to explain this question, so I'll give you them both.

WAY 1: First off, ignore the shaded box and just look at the other four. How many different orders are there for those four boxes? Well, there are 4 boxes that can go in 4 different spots, so that's 4[nPr]4, which is 24. (I'm totally ignoring the shaded box for now, hence there only being 4 spots instead of 5.)

So there are 24 different orders for the 4 non-shaded boxes. (ABCD, ABDC, ACBD, ACDB, ADBC, ADCB, etc...) For EACH of those orders, the shaded box can either be placed into the SECOND, THIRD, or FOURTH position (but not the first or fifth, according to the rules of the questions). So if X is the shaded box, you could have AXBCD, ABXCD, ABCXD... AXBDC, ABXDC, ABDXC... etc... So for EACH order, the shaded box could go in one of THREE spots. So that's 24 times 3, which is 72.

WAY 2: List out the options for each spot and multiply them out. There are five spots, so we have: __ __ __ __ __. The shaded box can't be the first or last, so for the FIRST spot, we have 4 options:
4 __ __ __ __. Once one card goes in the first spot, that means that there are 3 cards still available for the LAST spot (since the shaded card can't go there): 4 __ __ __ 3. Now we have three cards left (including the shaded card), and no more restrictions. The SECOND spot could be any of the 3 remaining cards, which would leave 2 for the THIRD spot, and 1 for the FOURTH spot: 4 3 2 1 3. Multiply them out and you get 72.