Question

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✔How many words can be formed by the letters of the word PERMUTATIONS if --

》the words begin with P and end with S ?

》The vowels always occur together ?

》 There be always 4 letters between P and S ?


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Answer 1

PERMUTATIONS

Total letters between P and S

= 10

2 T's are there

So No.of ways to arrange words that begin with P and end with S

= 10!/2

b)vowels occur together

PERMUTATIONS

EUAIO these are vowels

Left - PRMTTNS

total letters = 7 ( consonants) + 1( all vowels take as 1)

= 8

No of ways to arrange it

= 8!/2 ( as 2 T's are there)

Vowels arrangement = 5!

So No of ways when vowels occur together = 8! × 5! /2

c) 4 letters between P and S

PERMUTATIONS

ERMUTATION

total letters between P and S -10

Choose 4 letters and arrange - 10P4 = 10!/6!

2 T's - Divide by 2

10!/6! × 2

Between P and S ,here S and P can be interchanged

So multiply by 2

2× 10!/6! × 2

10!/6!

Now total 6 letters we got,rest 6 letters and 1 ( P (4 letters) S) can be arranged in 7! Ways

So 10! × 7!/6! = 10! ×7

Answer 2

Heya!

⚫the words begin with P and end with S ?

ERMUTATION = 10 WORDS

⚫There be always 4 letters between P and S ?

euai (any four)

hope it helps you