Find the number of arrangements of the letters of the word INDEPENDENCE
Find the number of arrangements of the letters of the word, INDEPENDENCE if all E are together.
Show Answer (Detailed Solution Below)Option 3 : 30240 Free Electric charges and coulomb's law (Basic) 10 Questions 10 Marks 10 Mins Concept: Let there be n things of which p1 are alike of one kind, p2 are alike of another kind, p3 are alike of 3rd kind, ..…, pr are alike of rth kind such that p1 + p2 + ….+ pr = n. Then the permutations of n objects is \(\rm \frac{n!}{(p_1!)\times (p_2!)\times ....\times (p_r!)} \) Calculation: The word INDEPENDENCE contains 12 letters out of which N occurs 3 times, D occurs 2 times, E occurs 4 times, and the rest of the letters occur only once. If E is together so all E count as 1 Then Total word is 9 Hence, required number of arrangements \(=\rm \frac{9!}{(3)! \times(2!)}\) \(=\rm \frac{9\times8\times7\times6\times5\times4\times3!}{(3!)\times(2)}\) \(=\rm9\times8\times7\times6\times5\times 2\) = 30240 Hence, option (3) is correct. Last updated on Sep 29, 2022 Union Public Service Commission (UPSC) has released the NDA Result II 2022 (Name Wise List) for the exam that was held on 4th September 2022. Earlier, the roll number wise list was released by the board. A total number of 400 vacancies will be filled for the UPSC NDA II 2022 exam. The selection process for the exam includes a Written Exam and SSB Interview. Candidates who get successful selection under UPSC NDA II will get a salary range between Rs. 15,600 to Rs. 39,100. There are 3N,4E,2D,1l,1P,1C Since letters are repeating so we use the formula Total number of arrangements are:
= 1663200 (i) If word starts with P. Now P will be fixed, We now need to arrange 11 letters. Where there are 4E,3N,2D Since letters are repeating so we use the formula
No. of arrangements = 138600 (ii) do all the vowels always occur together We will consider vowels as a same letter. Here there are 5vowels i.e IEEEE, So, these can be arranged in Now total letters are 7+1 = 8 Now there are 3N,2D in remaining letters. These can be arranged in
Total ways = 16800 iii) do all the vowels never occur together Vowels never occur together = Total arrangements – vowels occur together = 1663200 – 16800 = 1646400 iv) do the words begin with I and end in P? Let’s fix I and P at the ends. There are 2D,4E,3N. As letters are repeating, Total arrangements
= 12600 How many arrangements can be made from the word independence?Hence, the total number of arrangements made by rearranging the letters in the word independence are 1663200.
Which vowel is repeated in the word independence?To find in how many of these arrangements do all vowels occur together: Also, there are 5 vowels in the word INDEPENDENCE (E is repeated four times and I is repeated one time). If we want all vowels to be together then treat them as a single object.
What is the total number of possible 4 letter arrangements?The answer is 4! = 24. The first space can be filled by any one of the four letters.
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