is the answer of b) IV) 40?
5P2 for the 2 middle letter times 2P2 for the first and last letters
I've been stuck with the last two problems of this question using the fundemental counting principle.
Determine the number of distiguishable four letter arrangements that can be formed from the word Germany if:
a) Letters can be repeated=2401
b) no letters repeated and:
I) there are no further restrictions =840
II) the first letter must be M = 120
III) the "word" must cointain G=?
IV) the fist and last letters must be vowels=?
Thanks for any help
Hello, nightrider456!
Determine the number of distiguishable 4-letter arrangements
that can be formed from the word GERMANY if:
III) the "word" must cointain G.
"G" is already included.
Select 3 of the other 6 letters: . choices.
Now arrange the four letters: . ways.
Therefore, there are: . such arrangements.
I will assume that Y is not a vowel.IV) the first and last letters must be vowels.
We have two vowels {A, E} and five consonants {G, M, N, R, Y}
There are 2 choices for positioning the vowels: .A _ _ E . or . E _ _ A
Select 2 of the 5 consonants and place them in the middle blanks: . ways.
Therefore, there are: . such arrangements.