Question

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?

Answer 1

(i)

The word PERMUTATIONS contains 12 letters where T appears 2 times. If the words start with P and end with S, then this means that 2 positions in 12 letters word are already fixed. Then there are only 10 positions left with 10 letters that contains two T. Thus the permutation becomes,

10! 2!

Cancel the common factors from numerator and denominator by factorizing the greater term in form of factorials.

The formula to calculate the factors of a factorial in terms of factorial itself is,

n!=n( n−1 )! n!=n( n−1 )( n−2 )!                 [ n≥2 ] ⋮

The permutation becomes,

10! 2! = 10×9×8×7×6×5×4×3×2! 2! =1814400

Thus, the number of ways of arranging the letters is1814400.

(ii)

The word PERMUTATIONS contains 5 vowels E, U, A, I and O. Since they will occur together, thus they will be considered as one letter. Then the total letters in the word becomes8 (= 12 – 5 + 1, since 5 vowels combined into one). Out of this there are 2 T. So the number of possible permutations is,

8! 2!

The formula to calculate the factors of a factorial in terms of factorial itself is,

n!=n( n−1 )! n!=n( n−1 )( n−2 )!                 [ n≥2 ] ⋮

The permutation becomes,

8! 2! = 8×7×6×5×4×3×2! 2! =20160

Now, the 5 vowels which are taken together can be arranged in many different ways amongst each other. The number of permutations of 5 different vowels taken all at a time is given by:

P 5 5 = 5! ( 5−5 )! = 5! 0! = 5! 1 =5!

The formula to calculate n! is defined as,

n!=1×2×3×⋯×( n−1 )×n

The factorial of 5 is,

P 5 5 =5×4×3×2×1 =120

By multiplication principle which states that if an event can occur in m different ways and follows another event that can occur in n different ways, then total number of ways is m×n.

Therefore, the number of ways a word can be written with all vowels together is,

20160×120=2419200

Thus, required number of words in which vowels come together is 2419200.

(iii)

The letters have to be arranged in such a way that there are always 4 letters between P and S. Therefore, P and S are 4 letters apart. The number of ways of doing this is shown below:

P __ __ __ __ S__ __ __ __ __ __

__ P __ __ __ __ S __ __ __ __ __

__ __P __ __ __ __S __ __ __ __

…

…

__ __ __ __ __ __ P __ __ __ __ S

So there are 7 such positions. But P and S can also interchange their positions. So total number of permutations of P and S = 7 x 2 = 14

Now, out of 12 places, 2 are taken by P and S. There remains 10 letters with 2 T’s which have to be arranged in 10 places. The permutation becomes,

10! 2!

Cancel the common factors from numerator and denominator by factorizing the greater term in form of factorials.

The formula to calculate the factors of a factorial in terms of factorial itself is,

n!=n( n−1 )! n!=n( n−1 )( n−2 )!                 [ n≥2 ] ⋮

The permutation becomes,

10! 2! = 10×9×8×7×6×5×4×3×2! 2! =1814400

By multiplication principle which states that if an event can occur in m different ways and follows another event that can occur in n different ways, then total number of ways is m×n.

The total number of arrangement is,

1814400×14=25401600

Thus, required number of arrangements is 25401600 such that 4 letters comes between P and S.