Question

Question 11 In how many ways can the letters of the word PERMUTATIONS be arranged if the

(i) words start with P and end with S, (ii) vowels are all together,

(ii) there are always 4 letters between P and S?

Class X1 - Maths -Permutations and Combinations Page 148

Answer 1

Word ' PERMUTATIONS has
P - 1 time
E - 1 time
R - 1 time
M - 1 time
U - 1 time
T - 2 times
A - 1 time
I - 1 time
O - 1 time
N - 1 time
S - 1 times

(1) word start with P and end with S
P, —, —, —, —, —, —, —, —, —, —, S
you can see that 1st and last portion are already filled by P and S respectively .
now, remaining 10 portion can be filled = 10!/2! { because T is 2 times }
= 10× 9 × 8 × 7 × 6 × 5 × 4 × 3
= 1814400

(ii) vowels are all together .
if vowels are all together then, it will be considered as single letter . here all vowels are present { A , E , I , O , U} e.g 5 letters like as single letter .
now, rest 7 letters and 1 single letter in of vowels will be considered as 8 letters .
hence,
8 letter can be arranged = 8! ways
5 vowels can be arranged = 5! ways
hence , require number of PERMUTATIONS= 8! × 5!/2! { because T comes 2 times }
= 8 × 7 × 6 × 120 × 120/2
= 2419200


(iii) there are always 4 letters between P and S.

total number of letters to be arranged = 12
there are always 4 letters between P and S .
1st way => P, —, —,—,—,S,—,—,—,—,—,—
2nd way=> —, P,—,—,—,—,S,—,—,—,—,—
3rd way =>—,—,P,—,—,—,—,S,—,—,—,—
4th way=>—,—,—,P,—,—,—,—,S,—,—,—
5th way =>—,—,—,—,P,—,—,—,—,S,—,—
6th way =>—,—,—,—,—,P,—,—,—,—,S,—
7th way => —,—,—,—,—,—,P,—,—,—,—,S
here you can see that P and S having 4 places between them can be filled in 7 ways .
hence, P and S or S and P can be filled in (7+7) ways .
remaining 10 letters can be filled in 10!/2! ways

hence, fundamental principle of counting,
total number of ways = 14 × 10!/2!
= 25401600