In given word OLIVER there are 3 vowels E, I & O. These can be arranged in only one way as dictionary order E, I & O.
There are 6 letters in thegiven word.
First arrange 3 vowels.
This can be done in 6C3 ways and that too in only one way.(dictionary order E, I & O)
Remaining 3 letters can be placed in 3 places = 3! ways
Total number of possible ways of arranging letters of OLIVER = 3! x ways = 6x5x4 = 120 ways.
Choose 1 person for the single room & from the remaining choose 2 for the double room & from the remaining choose 4 people for the four person room,
Then, 7C1 x 6C2 x 4C4
= 7 x 15 x 1 = 105
From 5 consonants, 3 consonants can be selected in ways.
From 4 vowels, 2 vowels can be selected in ways.
Now with every selection, number of ways of arranging 5 letters is ways.
Total number of words =
= 10x 6 x 5 x 4 x 3 x 2 x 1= 7200
There are 7 letters in the word Bengali of these 3 are vowels and 4 consonants.
There are 4 odd places and 3 even places. 3 vowels can occupy 4 odd places in ways and 4 constants can be arranged in ways.
Number of words = x = 24 x 24 = 576
Here the order of choosing the elements doesn?t matter and this is a problem in combinations.
We have to find the number of ways of choosing 4 elements of this set which has 11 elements.
This can be done in 11 ways = 330 ways
Given word is THERAPY.
Number of letters in the given word = 7
Number of vowels in the given word = 2 = A & E
Required number of different ways, the letters of the word THERAPY arranged such that vowels always come together is
6! x 2! = 720 x 2 = 1440.
Number of words with 5 letters from given 9 alphabets formed =
Number of words with 5 letters from given 9 alphabets formed such that no letter is repeated is =
Number of words can be formed which have at least one letter repeated =
= 59049 - 15120
= 43929
ST candidates vacancies can be filled by ways = 10
Remaining vacancies are 5 that are to be filled by 12
=> = (12x11x10x9x8)/(5x4x3x2x1) = 792
Total number of filling the vacancies = 10 x 792 = 7920
The students should sit in between two teachers. There are 7 gaps in between teachers when they sit in a roundtable. This can be done in ways. 7 teachers can sit in (7-1)! ways.
Required no.of ways is =
Number of ways of choosing 2 black pens from 5 black pens in ways.
Number of ways of choosing 2 white pens from 3 white pens in ways.
Number of ways of choosing 2 red pens from 4 red pens in ways.
By the Counting Principle, 2 black pens, 2 white pens, and 2 red pens can be chosen in 10 x 3 x 6 =180 ways.
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