How many ways can Leading be arranged in such a way that atleast two vowels always together?

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Permutations and Combinations:

1. The number of ways to arrange 'r' objects out of 'n' distinct objects is expressed as: nPr=n! (n-r)!
2. If from the total set of n numbers, p is of one kind and q, r are others, respectively, then nPr=n!p!×q!×r! 
3. nPn=n!
4. The number of ways to arrange 'n' distinct letters in a row =n!
5. The number of ways to arrange 'n' objects in a row such that 'p' objects out of 'n' objects are identical and 'q' objects out of 'n' objects are identical is n!p!q!.
6. Sum of all the numbers formed by the arrangements of 'n' non-zero digits =(n-1)!Sum of all digits in the arrangement ×1111...n times.
7. Total number of circular permutations of 'n' objects, if the order of the circular arrangement (clockwise or anti-clockwise) is considerable, is =(n-1)!
8. Total number of circular permutations of 'n' objects, if there is 'No difference' between clockwise and anticlockwise arrangements, is =(n-1)!2.
9. nCr=nPrr!=n! (n-r)!r!
10. Number of ways to select 'zero or more' objects out of 'n' distinct objects =2n
11. Number of ways to select 'at least one' object out of 'n' distinct objects =2n-1
12. Number of ways to select 'r' objects out of 'n' identical objects =1
13. Number of ways to select 'zero or more' objects out of 'n' identical objects =n+1
14. Number of ways to select 'zero or more' objects out of 'p+q+r' objects of which p objects are alike of one kind, q objects are alike of second kind and r objects are alike of third kind =(p+1)(q+1)(r+1)
15. Number of ways to select 'at least one' objects out of 'p+q+r' objects of which p objects are alike of one kind, q objects are alike of second kind and r objects are alike of third kind is (p+1)(q+1)(r+1)-1.
16. Number of ways to select 'zero or more' objects out of 'p+q+r' objects of which p objects are alike of one kind, q objects are alike of second kind and r distinct objects =(p+1)(q+1)2r
17. Number of ways to select 'at least one' objects out of 'p+q+r' objects of which p objects are alike of one kind, q objects are alike of second kind and r distinct objects =(p+1)(q+1)2r-1
18. Number of selections of 'k' consecutive (adjacent) objects from a row of 'n' objects =(n-k+1)
19. Number of diagonals in an 'n' sided polygon =nC2-n

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1.

In how many different ways can the letters of the word 'LEADING' be arranged in such a way that the vowels always come together?

A. 360 B. 480 C. 720 D. 5040 Answer: Option C Explanation: The word 'LEADING' has 7 different letters. When the vowels EAI are always together, they can be supposed to form one letter. Then, we have to arrange the letters LNDG (EAI). Now, 5 (4 + 1 = 5) letters can be arranged in 5! = 120 ways. The vowels (EAI) can be arranged among themselves in 3! = 6 ways. Required number of ways = (120 x 6) = 720.

2.	From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?
A. 564 B. 645 C. 735 D. 756 Answer: Option D Explanation: We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only). Required number of ways = (7C3 x 6C2) + (7C4 x 6C1) + (7C5) = 7 x 6 x 5 x 6 x 5 + (7C3 x 6C1) + (7C2) 3 x 2 x 1 2 x 1 = 525 + 7 x 6 x 5 x 6 + 7 x 6 3 x 2 x 1 2 x 1 = (525 + 210 + 21) = 756.

3.

In how many different ways can the letters of the word 'CORPORATION' be arranged so that the vowels always come together?

A. 810 B. 1440 C. 2880 D. 50400 Answer: Option D Explanation: In the word 'CORPORATION', we treat the vowels OOAIO as one letter. Thus, we have CRPRTN (OOAIO). This has 7 (6 + 1) letters of which R occurs 2 times and the rest are different. Number of ways arranging these letters = 7! = 2520. 2! Now, 5 vowels in which O occurs 3 times and the rest are different, can be arranged Required number of ways = (2520 x 20) = 50400.

4.	In how many ways can the letters of the word 'LEADER' be arranged?
A. 72 B. 144 C. 360 D. 720 Answer: Option C Explanation: The word 'LEADER' contains 6 letters, namely 1L, 2E, 1A, 1D and 1R. Required number of ways = 6! = 360. (1!)(2!)(1!)(1!)(1!)

5.

Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?

A. 210 B. 1050 C. 25200 D. 21400 Answer: Option C Explanation: Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4)       = (7C3 x 4C2) = 7 x 6 x 5 x 4 x 3 3 x 2 x 1 2 x 1 = 210. Number of groups, each having 3 consonants and 2 vowels = 210. Each group contains 5 letters. Number of ways of arranging 5 letters among themselves = 5! = 5 x 4 x 3 x 2 x 1 = 120. Required number of ways = (210 x 120) = 25200.