Question

Find the arrangements of letters of the word INDEPENDENCE. In how many of these arrangements,

1. do the words start with P. 

2. do all the vowels always occur together. 

3. do the vowels never occur together. 

4. do the words begin with I and end in P?

Answer 1

In word INDEPENDENCE,

3Ns, 4Es, 2Ds and 1I, 1P and 1C are there (repetition)

n = 12, P₁ = 3, P₂ = 4 and P₃ = 2

∴ Total arrangements = \(\frac{n!}{P_1!P_2!P_3!}\)

\(= \frac{12!}{3!4!2!}\)

\(= \frac{12\times 11\times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 !}{3\times 2 \times 1 \times 4 ! \times 2 \times 1}\)

= 1663200

(i) words start with P

P−−−−−−−−−

So, 4Es, 3Ns, 2Ds (repetition) are there and n = 11 letters

∴ number of arrangements = \(\frac{11!}{4!3!2!}\)

\(= 138600\)

(ii) Arranging vowels.

Total number of vowels (n) = 5

Repetition → 4Es and 1I

∴ n = 5 and P₁ = 4

Total arrangement = \(\frac{5!}{4!}\)

Arranging the letters when vowels are taken as one unit.

Total number = n = 8

Here, 3Ns, 2Ds → repetition

Total arrangements = \(\frac{8!}{3!2!}\)

∴ Required number of arrangements = \(\frac {8!}{3! \times 2!} \times \frac{5!}{4!}\) = 16800

(iii) Number of arrangements when vowels never occur together = total number of arrangements (without any restriction) − the number of arrangements where all the vowels occur together

= 1663200 − 16800

= 1646400

(iv) Let I and P fix at extreme ends.

I−−−−−−−−−−P

10 letters in which 2D, 4E and 3N → repetition

So, n = 10, P₁ = 2, P₂ = 4 and P₃ = 3

∴ Required number of arrangements = \(\frac{10!}{2!4!3!}\) = 12600.

Answer 2

1. In the word INDEPENDENCE there are 12 letters, of which N appears 3 times, E appears 4 times, D appears 2 times and the rest all are different.

When the words start with P, then there are 11 letters to be filled in 11 spaces. Therefore the total number of ways is \(\frac{11 !}{3! \times 2 ! \times 4 !}\) = 138600.

2. The vowels EEEEI are to be kept together and should be treated as one unit. Then these vowels can be arranged in \(\frac{5 !}{4 !}\) ways. This single unit together with 7 letter will count to units, can be arranged in \(\frac{8 !}{3! \times 2 !}\). Therefore the total number of ways \(\frac{5 !}{4 !} \times \frac{8 !}{3 ! \times 2 !}\) = 16800.

3. Number of ways of arrangement with vowels do not come together = Total arrangement – vowels coming together.
= \(\frac{12 !}{3 ! \times 2 ! \times 4 !}\) – 16800 = 1663200 – 16800 = 1646400.

4. When the words start with I and ends with P, then there are 10 letters to be filled in 10 spaces. Therefore the total number of ways is \(\frac{10 !}{3 ! \times 2 ! \times 4 !}\) = 12600.