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A B C D E

If the 5 cards labled A throug E are placed in a row so that C is never at either end, how many different arrangements are possible?

By Tigeruppercut (Tigeruppercut) on Monday, September 20, 2004 - 03:44 am: Edit

a,b,c,d,e, = 5 choices
5x4x3x2x1 = if c could be placed at end
5! = 120

then removing two choices since c cannot be placed at end1 and end2: 120 - 2 = 118?

wtf i'm baffled too. tell me if i'm wrong or right.

By Glowingamy (Glowingamy) on Monday, September 20, 2004 - 03:50 am: Edit

3x4x3x2x1 = 72

I ASSUME NO RESPONSIBILITY IF YOU ARE EXPELLED FOR STUPID ANSWERS

By Imblue (Imblue) on Monday, September 20, 2004 - 06:40 am: Edit

Alternatively:

5! = 120 arrangements that allow C at either end

2 x 4! = 48 arrangements that have C at either end

answer = 120 - 48 = 72

By Drusba (Drusba) on Monday, September 20, 2004 - 09:30 am: Edit

72 is correct. Explanation: you start with a permutation of 5 taken 5 at a time for which the total is 5! or 120. When you say C cannot be at the beginning, what you are actually doing is removing a permutation of 4 items (ABDE) that would follow the C at the beginning and thus remove 4! (24). You then have to remove the same amount when C is at the end.

By Tigeruppercut (Tigeruppercut) on Monday, September 20, 2004 - 10:30 pm: Edit

drusba, can u explain in greater detail? i still don't get it

By Glowingamy (Glowingamy) on Tuesday, September 21, 2004 - 03:31 am: Edit

Er I got mine with dumb logic so maybe it will be easier to understand. You can see why I almost failed this test last year

1) There are 3 max. choices for the far left card. so 3, for 3 choices
3x

2) There WOULD be 5 card choices here, but you used one up for space #1 so there are 4
3x4x

3) again there would be five but you've used up two choices so there are 3
3x4x3

4) would be five, you used up 3 so there are 2
3x4x3x2

5) would be three, but whatever since there's only one left anyway!
3x4x3x2x1 = 72

By Imblue (Imblue) on Tuesday, September 21, 2004 - 04:31 am: Edit

If you have n objects, then there are n! different arrangements of these objects (permutations), where n! = n(n-1)(n-2)...1. This is because you have n possible objects to pick for the first position, n-1 for the second, ..., and 1 object left to put in the last position. So 5 objects have 5! = 5x4x3x2x1 = 120 different arrangements. Now let's find how many of these arrangements have C at either end, so we can subtract that number to get our answer. Suppose C is at the beginning. By the previous explanation, there are 4! = 4x3x2x1 = 24 such arrangements. The same goes for arrangements with C at the end. So your answer is 5! - 2x4! = 72.

By Adadamongdads (Adadamongdads) on Tuesday, September 21, 2004 - 10:02 am: Edit

Try the Art of Problem Solving - a math dedicated website.

By Tigeruppercut (Tigeruppercut) on Tuesday, September 21, 2004 - 10:59 pm: Edit

glowingamy, imblue. thanks a lot. i'm understanding it much better now.

By Tigeruppercut (Tigeruppercut) on Tuesday, September 21, 2004 - 11:08 pm: Edit

adadamongdads,

i just went ahead and bought the book.
looks great

thanks