Class 11th

22. There are 12 letters in which T appears 2 times and rest are all different.

i. When P and S are fixed as first and last letter we can arrange the remaining 10 letter taking all at a time. i.e.

Number of permutation = $\frac{10!}{2!}$

= 18,14,400

ii. We take the 5 vowels (E, U, A, I, O) as one single object. This single object with the remaining 7 object are treated as 8 object which have 2 – T's.

So, number of permutations in which the vowels come together

= permutation of 8 object x permutation within the vowels

= $\frac{8!}{2!}$ * 5!

= 20160 * 120

= 2419200

iii. In order to have 4 letters between P and S, (P, S) should have the possible sets of places (1, 6), (2, 7), (3, 8), (4, 9), (5, 10), (6, 11) and (7, 12) similarly (S, P) should have the same possible sets of places.

So, total number of possible positions fixed for P and S = 14

After fixing P and S we can rearrange the remaining 10 places with the remaining 10 letters of which 2 are T's.

So, required permutation = $\frac{10!}{2!}$

= 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3

= 18,14,400

Therefore, total permutation for which there are always 4 letters between P and S

= 14 * 18,14,400

= 2,54,01,600

20. i. The permutation of 6 letters in MONDAY taken 4 at a time without repetition is

ii. The permutation of 6 letters in MONDAY when all letters are taken at a time is

⁶P₆ = $\frac{6!}{\left(6-6\right)!}$ = $\frac{6!}{0!}$ = 6! = 6 * 5 * 4 * 3 * 2 * 1 = 720

iii. The permutation of having one of the two vowels (O, A) as first letter from the word MONDAY when all letters are taken at a time is

²P₁ = $\frac{2!}{\left(2-1\right)!}$ = 2

After fixing one of the vowel as first letter we can rearrange the remaining 5 letters taking 5 at a time

⁵P₅ = $\frac{5!}{\left(5-5\right)!}$ = $\frac{5!}{0!}$ = 5! = 5 * 4 * 3 * 2 * 1 = 120

Therefore, total permutation when all letters are used but first letter is vowel from the word MONDAY = 2 * 120 = 240