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continuing this process to z=24, we get: | continuing this process to z=24, we get: | ||
| − | <math>\left ( 25 \times 24 + 24 \times 23 + 23 \times 22 + ... + 2 \times 1 \right ) \times \frac {23!}{26!} = \frac{5200}{26 \times 25 \times 24} = \frac {1}{3} </math> | + | <math>\left ( 25 \times 24 + 24 \times 23 + 23 \times 22 + ... + 2 \times 1 \right ) \times \frac {23!}{26!} = \frac{5200}{26 \times 25 \times 24} = \frac {1}{3} </math> -mkburges |
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Brandy, | Brandy, | ||
If you still want help on part a, I think it's similar to the alphabet problems we did before that were about counting the permutations that contain fish, bird and cat or whatever. Anyway, what I did was group the first 13 letters as one "object," then look for permutations of that and the other 13 letters. I got 14! / 26! for an answer. Hope this helps, -Zoe | If you still want help on part a, I think it's similar to the alphabet problems we did before that were about counting the permutations that contain fish, bird and cat or whatever. Anyway, what I did was group the first 13 letters as one "object," then look for permutations of that and the other 13 letters. I got 14! / 26! for an answer. Hope this helps, -Zoe | ||
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| + | -- I tried the long way but i kept getting 1/6 for my answer. Watch out that you realize a and b can be in either order. | ||
Revision as of 17:10, 25 February 2009
How do I start this problem? I'm drawing a complete blank..It seems very familiar like we've done something like this in class or something, but I can't figure it out. Please help. -Brandy
Cool thing about f):
The question calls for the probability that the letter z precedes both a and b in the permutation of the 26 letters of the alphabet. This is basically saying what is the probability that when you mix up all the letters of the alpahabet (a,b,c,..z) that z will come before a and b.
Now, we can solve this the long way (later), or use your head a little: for any permutation of the alphabet, we can circle the letters a, b, and z. These can be arrange in 6 different ways if we just move these three around keeping all others the same. Therefore, the amount of letters we have between a, b, and z don't matter. We can have, 1, 100, or 0. Therefore the question becomes what is the probability that z comes first out of a, b, z. Which the answer is 1/3.
Now, if you don't believe me, here's the long way: we count how many ways that z can be in front of a and b:
if z=1, then a=25 ways and b=24 ways, and 23! others (in which to order them)
if z=2, then a=24 ways and b=23 ways, and 22! others
...
continuing this process to z=24, we get:
$ \left ( 25 \times 24 + 24 \times 23 + 23 \times 22 + ... + 2 \times 1 \right ) \times \frac {23!}{26!} = \frac{5200}{26 \times 25 \times 24} = \frac {1}{3} $ -mkburges
I did part f a little differently. First, do 26C3 in order to choose the 3 spots that z,a, and b (in that order) will be in. Multiply this number by 2 because for every combination of 3 spots we choose we can switch the order of a and b. Lastly, we have to find how many permutations there are of the 23 letters that are left. Then we divide this by the total number of permutations of all 26 letters. So, we have [2(26C3)*(23P23)]/(26P26)]=1/3. I hope that was clear enough. -user:cwithey
Brandy, If you still want help on part a, I think it's similar to the alphabet problems we did before that were about counting the permutations that contain fish, bird and cat or whatever. Anyway, what I did was group the first 13 letters as one "object," then look for permutations of that and the other 13 letters. I got 14! / 26! for an answer. Hope this helps, -Zoe
-- I tried the long way but i kept getting 1/6 for my answer. Watch out that you realize a and b can be in either order.
