Total number of words = 5! = 120
combining the vowels at one place(OEA) with remaining 2 letters MG, letters can be arranged in 3! ways. Also, three vowels can be arranged in 3! ways
So, when vowels are together, then number of words = 3! x 3! = 36
there4; Required number of ways, when vowels being never together =120 - 36 = 84
Three vowels (O, E, A) can be arranged in the odd places in 3! ways (1st position, 3rd position, 5th position) and two consonant (M, G) can be arranged in the even place in 2! ways (2nd place and 4th place).
? Total number of ways = 3! X 2! = 12
When E is fixed in the middle, then there are four places left to be filled by four remaining letters O, M, G and A and this can be done in 4! ways.
? Total number of ways = 4! = 24
When O and A occupy end places. Then, the three letters (M, E, G) can be arranged themselves by 3! = 6 ways and two letters (O, A) can be arranged among themselves in 2! = 2 ways.
? Total number of ways = 6 x 2 = 12
Number of ways in which 8 persons can be selected from 15 persons = 15C8
Now, 8 persons can be seated around a circular table in 7! ways.
Now, remaining 7 persons can be seated around a circular table in 6! ways.
? Required number of ways = 15C8 x 7! x 6!
Required number of ways
= 2n - 1 = 210 - 1 = 1024 - 1 = 1023
Total number of letters = 8
Number of vowels = 3 and r occurs twice.
Total number of arrangements when there is no restriction = 8!/2!
When three vowels are together, regarding them as one letter, we have only 5 + 1 = 6 letters
These six letters can be arranged in 6!/2! ways, since r occurs twice.
But the three vowels can be arranged among themselves in 3! ways.
Hence number of arrangement when the three vowels are together = 6! /(2 !x 3!)
? Required number = 8!/2! - {6! / (2! x 3!)} = 18,000
Married couples :
MF MF MF MF ? AB, CD, EF, CD
Possible teams :
AD CB EB GB
AF CF ED GD
AH CH EH GF
Team AD can play only with CB, CF, CH, EB, EH, GB, GF (7 teams).
Teams AD cannot play with AF, AH, ED and GD.
The same will apply with all teams, So, number of total matches = 12 x 7 = 84
But every match includes 2 teams, so the actual number of matches = 84/2 = 48
There is a 7 - digit telephone number but extreme right and extreme left positions are fixed .
i. e. , 6 x x x x x 5
? Required number of ways = 8 x 7 x 6 x 5 x 4 = 6720
First , we select 13 persons out of 24 persons in 24 C13 ways.
Now, these 13 persons can be seated in 12! ways around a table .
So required number of ways = 24 C13 x 12!
= [24! / {13!(24 - 13)!}] x 12!
= [24! / {13! x 11!}] x 12!
= 24! / (13 x 11!)
The required number of triangles = nC3 - mC3
Here, n = 14, m = 4
= 14C3 - 4C3
= (14 x 13 x 12 x 11! ) / (3! x 11!) - 4! / (3! x 1!)
= (14 x 13 x 12)/6 - 4/1
= 14 x 26 - 4
= 364 - 4
= 360
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