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I'm a little confused about part a. I tried it in two different ways that both seemed right to me and got very different answers. I figured to count all 6-letter strings containing ''a'', you could count all possible 6-letter strings and subtract the ones that do NOT contain ''a''. This should be <math>26^6-25^6=64,775,151</math>. | I'm a little confused about part a. I tried it in two different ways that both seemed right to me and got very different answers. I figured to count all 6-letter strings containing ''a'', you could count all possible 6-letter strings and subtract the ones that do NOT contain ''a''. This should be <math>26^6-25^6=64,775,151</math>. | ||
Then, I thought to try it this way: suppose I first pick any 5 letters (in <math>26^5</math> ways). Then, to guarantee the string contains an ''a'', I can insert an ''a'' into this string at any of 6 places. So with this method, my answer is <math>(26^5)*6=71,288,256</math>. What am I doing wrong here? | Then, I thought to try it this way: suppose I first pick any 5 letters (in <math>26^5</math> ways). Then, to guarantee the string contains an ''a'', I can insert an ''a'' into this string at any of 6 places. So with this method, my answer is <math>(26^5)*6=71,288,256</math>. What am I doing wrong here? | ||
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| + | UPDATE: I figured out that the second method is definitely wrong because it overcounts. For example, I could guarantee an ''a'' as the first character and have an ''a'' randomly as the second character. I could also guarantee the ''a'' in the second position and have one appear randomly in the first position. These two strings would be indistinguishable but counted multiple times. I think the first one is right (<math>26^6-25^6=64,775,151</math>). | ||
Revision as of 12:43, 14 September 2008
Part A
I'm a little confused about part a. I tried it in two different ways that both seemed right to me and got very different answers. I figured to count all 6-letter strings containing a, you could count all possible 6-letter strings and subtract the ones that do NOT contain a. This should be $ 26^6-25^6=64,775,151 $. Then, I thought to try it this way: suppose I first pick any 5 letters (in $ 26^5 $ ways). Then, to guarantee the string contains an a, I can insert an a into this string at any of 6 places. So with this method, my answer is $ (26^5)*6=71,288,256 $. What am I doing wrong here?
UPDATE: I figured out that the second method is definitely wrong because it overcounts. For example, I could guarantee an a as the first character and have an a randomly as the second character. I could also guarantee the a in the second position and have one appear randomly in the first position. These two strings would be indistinguishable but counted multiple times. I think the first one is right ($ 26^6-25^6=64,775,151 $).
