Consider “words” consisting of any string of 5 upper case letters...

fenixOG

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Consider “words” consisting of any string of 5 upper case letters...

Consider “words” consisting of any string of 5 upper case letters fromthe first half of the usual alphabet: A,B,C,D,E,F,G,H,I,J,K,L,M.
(a) How many words contain at least one A?
If 13^5 = all words we can make (371,293)
12^5 = all words ex. A's (248,832)

== 122,461


(b) How many words contain exactly two A’s?

would this be: 13 x (12 choose 3) x (5 choose two) x 3! ?
 
Consider “words” consisting of any string of 5 upper case letters from the first half of the usual alphabet: A, B, C, D, E, F, G, H, I, J, K, L, and M.

(a) How many words contain at least one A?


My answer:
If 13^5 = all words we can make (371,293)
12^5 = all words ex. A's (248,832)

== 122,461
I haven't checked your arithmetic, but your method looks fine. If any of the thirteen letters can fill any of the five slots, any number of times, then there are (13)(13)(13)(13)(13) = 13^5 ways of creating those five-letter "words". If one excludes the letter A from consideration, then there are only twelve options for each slot, so there are only (12)(12)(12)(12)(12) = 12^5 ways of creating those five-letter "words".

To say that a "word" contains "at least one A" is to say the opposite of "contains no A's at all". One method for finding this number of options (being "at least one") is to find the opposite (being "none at all") and subtracting from the full set (being 13^5, in this case).

So, assuming your arithmetic is correct, your answer should be, also.

(b) How many words contain exactly two A’s?

My answer:
Would this be: 13 x (12 choose 3) x (5 choose two) x 3! ?
How are you getting this answer? What was your reasoning? (You can explain your reasoning in a manner similar to what I showed for part (a) above.)

For any given five-letter "word" under these conditions, two of the letters are fixed as being A's; the other three could be any of the remaining twelve options. But the placement of those two A's is not fixed. So how many options would there be, if you consider the placement of those A's? And so forth.

Please be complete. Thank you! ;)
 
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