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<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> | ||
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| + | 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. So, we have 2(26C3)*(23P23)=1/3. I hope that was clear enough. -[[cwithey|user:cwithey]] | ||
Revision as of 07:59, 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} $
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. So, we have 2(26C3)*(23P23)=1/3. I hope that was clear enough. -user:cwithey
