How many different 6 letter "words" can be formed from the letters C,A,N,Y,O,N if:
O must be last, A must be first and repetitions aren't allowed. Here is how I set it up: A ? ? ? ? O (? Represents blanks) so 4!?
so you would have 4 choice at first, then 3 choices next, then 2, then 1
4*3*2*1=24 possible combinations
agree
To find the number of different 6-letter words that can be formed from the given letters with the given conditions, we can break down the problem step by step.
First, we know that the letter "A" must be first and the letter "O" must be last.
So let's set up the word format: A _ _ _ _ O
Now, we have 4 remaining letters to place in the 4 remaining blanks. Since repetitions are not allowed, every blank must be filled with a different letter from the remaining set of letters: C, N, and Y.
To fill the first blank, we have 3 options (C, N, and Y).
To fill the second blank, we have 2 options (the two remaining letters).
To fill the third blank, we have 1 option (the remaining letter).
So the number of different arrangements for the 4 remaining letters is: 3 * 2 * 1 = 6.
Therefore, the total number of different 6-letter words that can be formed is: 6 (from the 4 remaining letters) * 1 (since the position of "A" and "O" are fixed) = 6.
Hence, there are 6 different 6-letter words that can be formed from the letters C, A, N, Y, O, with the given conditions.