Question 1204819
<font color=black size=3>
Answers: 
(a) <font color=red>16807</font>
(b) <font color=red>2520</font>
(c) <font color=red>2401</font>
(d) <font color=red>720</font>



------------------------------------------------
------------------------------------------------


Explanation for part (a)


There are 7 letters in the set {A,B,C,D,E,F,G}


When no condition is imposed, we can repeat letters. 
There are 7 choices and 5 slots to fill. Giving 7^5 = <font color=red>16807</font> different words possible. 
Most of these "words" won't be found in the dictionary, but are simply strings of the letters mentioned. 


------------------------


Explanation for part (b)


Unlike the previous part, we cannot repeat letters here.
There are 7 choices for the first slot, then 6 for the next slot, and so on.


We have 7*6*5*4*3 = <font color=red>2520</font> different permutations possible.


Alternatively, use the nPr formula with n = 7 and r = 5.


------------------------


Explanation for part (c)


The first slot is locked to letter A.
The remaining 4 letters can be anything of the set {A,B,C,D,E,F,G} where repeats are allowed. 
The logic is similar to part (a) we've done earlier.
We have 7^4 = <font color=red>2401</font> different words that start with A when repeat letters are allowed.


------------------------


Explanation for part (d)


Given set = {A,B,C,D,E,F,G}
vowels from that set = {A, E}


There are 2 choices for the second slot (either A or E)
There are 7-1 = 6 choices for the first slot, due to one of the vowels already taken.
Then 6-1 = 5 choices for the third slot, 5-1 = 4 choices for the fourth slot, and 4-1 = 3 choices for the last slot.


6*2*5*4*3 = <font color=red>720</font> permutations where a vowel is the second letter.
</font>