How many permutations of the letters a, b, c, d, e, f, g have either two or three letters between a and b? b _ _ a is also very much possible.
nPr= n!/(n - r)!, where n >= r
The Attempt at a Solution
For this question there can be 4 cases which are as follows
1)when there are 4 letter words,
a _ _ b
from among 5 remaining letters 2 can be taken in 5P2 ways which can be arranged themselves in 2! ways and a and b can also be arranged among themselves in 2 ways, so
5P2*2!*2! = 80
2)5 letter words, here can be 3 cases too which are as follow:-
A) a _ _ b _
B) a _ _ _ b
c) _ a _ _ b
letters can be arranged here as
(5P3*3!*2)*3 = 2160.
3) when 6 lettered words are formed
a) a _ _ _ b _
b)a _ _ b _ _
c)_ a _ _ b _
d)_ a _ _ _ b
e) _ _ a _ _ b
here the letters can be arranged as
(5P4*4!*2)*5 = 28800
4)when 7 lettered word is formed
a) a _ _ _ b _ _
b) a _ _ b _ _ _
c) _ _ a _ _ _ b
d) _ _ _ a _ _ b
e) _ a _ _ b _ _
f) _ a _ _ _ b _
g) _ _ a _ _ b _
(5P5*5!*2)*7 = 201600
So now am getting the answer as 232640.
Please tell me am I right? If not then where am I making mistake.