What is correct arrangement of these letters INDIA *?

Example 16 Find the number of arrangements of the letters of the word INDEPENDENCE. In how many of these arrangements, Finding total number of arrangements In word INDEPENDENCE There are 3N, 4E, & 2D, 1I, 1P & 1C Since letters are repeating, so we use this formula 𝑛!/𝑝1!𝑝2!𝑝3! Total letters = 12 So, n = 12 Since, 3N, 4E, & 2D p1 = 3, p2 = 4,p3 = 2 Total arrangements = 12!/3!4!2! = 1663200 Example 16 Find the number of arrangements of the letters of the word INDEPENDENCE. In how many of these arrangements, do the words start with P If the word start with P We need to arrange (12 – 1) = 11 We need to arrange letters I, N, D, E, E, N, D, E, N, C, E Here, we have 4E, 3N,2D Since letters are repeating since we use this formula Number of arrangements = 𝑛!/𝑝1!𝑝2!𝑝3! Total letters to arrange = 11 So, n = 11 Since, 4E, 3N,2D p1 = 4 , p2 = 3 , p3 = 2 Number of arrangements = 𝑛!/𝑝1!𝑝2!𝑝3! = 11!/(4! 3! 2!) = 138600 Example 16 Find the number of arrangements of the letters of the word INDEPENDENCE. In how many of these arrangements, (ii) do all the vowels always occur together There are 5 vowels in the given word “INDEPENDENCE” i.e. 4E’s & I’s They have occur together we treat them as single object we treat as a single object So our letters become We arrange them now Arranging 5 vowels Since vowels are coming together, they can be and so on In EEEEI, there are 4E Since letters are repeating, We use this formula = 𝑛!/𝑝1!𝑝2!𝑝3! Total letter = n = 5 As 4E are there, p1 = 4 Total arrangements = 5!/4! Arranging remaining letters Numbers we need to arrange = 7 + 1 = 8 Here are 3N, 2D Since letter are repeating, We use this formula = 𝑛!/𝑝1!𝑝2!𝑝3! Total letters = n = 8 As 3N, 2D p1 = 3 , p2 = 2 Total arrangements = 8!/3!2! Hence, Required number of arrangements = 8!/(3! 2!) × 5!/4! = ((8 × 7 × 6 × 5 × 4 × 3!) × (5 × 4!))/(3! × 4! × 2) = 16800 Example 16 Find the number of arrangements of the letters of the word INDEPENDENCE. In how many of these arrangements, (iii) do the vowels never occur together Number of arrangements where vowel never occur together = Total number of arrangement – Number of arrangements when all vowels occur together = 1663200 – 16800 = 1646400 Example 16 Find the number of arrangements of the letters of the word INDEPENDENCE. In how many of these arrangements, (iv) do the words begin with I and end in P? Lets fix I and P at Extreme ends Since letters are repeating, Hence we use the formula 𝑛!/𝑝1!𝑝2!𝑝3! Here, Total letters = n = 10 Since 2D, 4E, 3N p1 = 2, p2 = 4, p3 = 3 Required number of arrangement = 10!/2!4!3! = 12600

Solution : The word 'INDIA' contains 2 I's, 1 A, 1 N and 1 D.
Number of permutations of the letters of the given word `=(5!)/(2!)=60.`
Let us assume that the vowels IIA form 1 letter.
Now, IIA + ND gives 3 letters, all district.
Number of their arrangements = 3! = 6.
Now, IIA has 3 letters, out of which there are 2 I's and 1 A.
Number of their arrangements `=(3!)/(2!)=3.`
Number of words having all vowels together `=(6xx3)=18.`
Number of words in which all vowels are not together `=(60-18)=42.`

Answer : (i) ADIIN (ii) DAIIN (iii) IADIN (iv) NADII (v) NIIDA

Solution : The word 'INDIA' contains 5 letters out of which there are 2 I's, 1 N, 1 D and 1 A.
Number of arrangements of its letters `=(5!)/(2!)=60.`
In the dictionary, the first word is ADIN.
Starting with A, we arrange the letters D, I, I, N in `(4!)/(2!)=12` ways.
The 13th word is DAIIN.
Sarting with D, we arrange A, I, I, N in `(4!)/(2!)=12` ways.
The 25th word is IADIN.
Starting with I, we arrange the letters A, D, I, N in 4! = 24 ways.
The 49th word is NADII.
Starting with N, we arrange the letters A, D, I, I in `(4!)/(2!)=12` ways.
The 60th word is NIIDA.

The letters of the word ‘INDIA’ are arranged as in a dictionary.

Question:

The letters of the word ‘INDIA’ are arranged as in a dictionary. What are the $1^{\text {st }}, 13^{\text {th }}, 49^{\text {th }}$ and $60^{\text {th }}$ words?

Solution:

Alphabetical arrangement of letters: A,D,I,N

$\Rightarrow 1^{\text {st }}$ word: ADIIN

To find other words:

Case 1: words starting with A

Number of words $=\frac{4 !}{2 !}=12$

$\Rightarrow 13^{\text {th }}$ word starts with D and is DAllN

Case 2: words starting with D

Number of words $=\frac{4 !}{2 !}=12$

Case 3: Words starting with I

Number of words $=4 !=24$

$\Rightarrow(12+12+24+1)^{\text {th }}=49^{\text {th }}$ word starts with $\mathrm{N}$ and is $\mathrm{NAllD}$

Note:The students must remember that to approach this problem, if they actually start writing all the words, it will become really difficult as there are a total of 60 words possible.The students must note that “permutation and combination” makes our tasks so easy. We cannot go on writing all the words and combinations possible out of this word. So, we have an easy way to find its number at least.

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