|By Mitodnaman (Mitodnaman) on Wednesday, June 02, 2004 - 03:56 pm: Edit|
If the 5 cards are placed in a row so that 1 card is never at either end, how many differnt arrangements are possible?
This is it for the present.
|By Crypto86 (Crypto86) on Wednesday, June 02, 2004 - 03:59 pm: Edit|
Your question doesn't make sense. It probably should be 5! (permutation arrangement of n items = n!) but "so that 1 card is never at either end" doesn't make sense. Is it that the "1" card is never at either end? The way you said it then there can never be any arrangments, because always there is some card at either end.
|By Optimizerdad (Optimizerdad) on Wednesday, June 02, 2004 - 04:16 pm: Edit|
Let's rephrase the question - if we have 5 cards labelled '1', '2', ..., '5', how many different arrangements are possible so that the '1' card is at neither end?
The answer is (total #arrangements of 5 cards) - (# arrangements with '1' card at beginning) - (# arrangements with '1' card at end)
= 5! -4! -4!
= (1)(2)(3)(4)(5) - (1)(2)(3)(4) -(1)(2)(3)(4)
= 120 - 24 - 24
|By Qwert271 (Qwert271) on Wednesday, June 02, 2004 - 04:28 pm: Edit|
An easier way to think about it is:
You have 4 choices for the #1 spot, 3 coices for the #5 spot, 3 choices for the #3 spot, 2 choices for the #2 spot, and 1 choice for the #4 spot = 4*3*3*2*1 = 72
|By Optimizerdad (Optimizerdad) on Wednesday, June 02, 2004 - 04:42 pm: Edit|
Nice trick, processing the spots in the sequence #1,#5,#3,#2,#4 . Trying them in the instinctive sequence #1,#2,#3,#4,#5 would've given you problems, I think!
|By Mitodnaman (Mitodnaman) on Thursday, June 03, 2004 - 06:14 pm: Edit|
how about these:
the 6 cabins at a camp are arranged so that there is exactly 1 stragiht path between each 2 of them, but no 3 of them are on a straight path. What is the total number of such straight paths joining these cabins?
The cells of a certain type of bacteria increase in number by each splitting into 2 new cells every 30 minutes. At that rate, if a colony of this type of bacteria starts with a single cell, how many hours elapse before the colony contains 2^11 cells?
I used total=pe^(rt) for this one and got some completely different answer.
|By Optimizerdad (Optimizerdad) on Thursday, June 03, 2004 - 07:52 pm: Edit|
Q1. Think of the cabins as being the points of a hexagon. There are 5 paths linking C1 to C2,C3,...,C6, then 4 paths linking C2 to C3,C4,..C6 etc. Answer is 5+4+3+2+1 = 15 paths.
Q2. After 1 half_hour period, you have 2^1 cells; after 2 half_hour periods, you have 2^2 cells, and so on. So, after 11 half_hour periods (or 5.5 hours) you have 2^11 cells.
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