Find the number of different 8-letter arrangements that can be made from the letters of the word EDUCATION so that all vowels do not occur together.(a) 40320(b) 37440(c) 1440(d) 2880I have been asked this question in examination.The origin of the question is Permutations-2 topic in division Permutations and Combinations of Mathematics – Class 11

Find the number of different 8-letter arrangements that can be made fr

1.	Find the number of different 8-letter arrangements that can be made from the letters of the word EDUCATION so that all vowels do not occur together.(a) 40320(b) 37440(c) 1440(d) 2880I have been asked this question in examination.The origin of the question is Permutations-2 topic in division Permutations and Combinations of Mathematics – Class 11
Answer» Correct CHOICE is (b) 37440 To explain: There are 5 VOWELS in word EDUCATION. 5 vowels can be arranged in ^5P5 i.e. 5! WAYS. When all vowels are TOGETHER, 5 vowels together form one letter and remaining 3 letters i.e. together 4 letters can be arranged in ^4P4 i.e. 4! ways. Total possible arrangements are 5! * 4! = 120*24 = 2880. So, when all vowels do not occur together, total possible arrangements = 8! – 2880 = 40320 – 2880 = 37440.

Discussion

No Comment Found

Related InterviewSolutions

^4Pr = 4*^5Pr-1. Find r.(a) 1(b) 2(c) 3(d) 4The question was posed to me during an online interview.My question is based upon Permutations-2 in division Permutations and Combinations of Mathematics – Class 11
Find the number of 5 letter words which can be formed from word IMAGE without repetition using permutations.(a) 20(b) 60(c) 120(d) 240The question was asked in class test.My doubt stems from Permutations-1 topic in section Permutations and Combinations of Mathematics – Class 11
The number of permutations of n different objects taken r at a time, where repetition is allowed is _______________(a) n!(b) r!(c) ^nPr(d) n^rThe question was asked in semester exam.This interesting question is from Permutations-1 in portion Permutations and Combinations of Mathematics – Class 11
A child has 2 pencil and 3 erasers. In how many ways he can take a pencil and an eraser?(a) 5(b) 6(c) 8(d) 9This question was addressed to me by my school principal while I was bunking the class.This is a very interesting question from Fundamental Principle of Counting topic in section Permutations and Combinations of Mathematics – Class 11
In a family, 5 males and 3 females are there. In how many ways we can select a group of 2 males and 2 females from the family?(a) 3(b) 10(c) 30(d) 40The question was asked by my college director while I was bunking the class.My query is from Combinations in chapter Permutations and Combinations of Mathematics – Class 11
Out of a group of 5 persons, find the number of ways of selecting 3 persons.(a) 1(b) 5(c) 10(d) 15I got this question in unit test.This interesting question is from Combinations in division Permutations and Combinations of Mathematics – Class 11
^nPr = ^nCr * ______________(a) r!(b) 1/r!(c) n!(d) 1/n!I had been asked this question by my school principal while I was bunking the class.My doubt stems from Combinations in portion Permutations and Combinations of Mathematics – Class 11
Order matters in combination.(a) True(b) FalseThe question was asked in unit test.The above asked question is from Combinations topic in division Permutations and Combinations of Mathematics – Class 11
If there are 4 paths to travel from Delhi to Kanpur, then in how many ways a person can travel from Delhi to Kanpur and came back to Delhi via same path?(a) 4(b) 8(c) 12(d) 16The question was posed to me in quiz.My question is taken from Fundamental Principle of Counting in portion Permutations and Combinations of Mathematics – Class 11
^nPn = ________________(a) n!(b) 1(c) \(\frac{1}{(n)!}\)(d) (n-1)!The question was posed to me in examination.My doubt is from Permutations-1 in portion Permutations and Combinations of Mathematics – Class 11

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment