There are 7 letters in the word DRAUGHT, the two vowels are A and U. Since, the vowels are not to be separated, AU can be considered as one entity. Therefore, the number of letters will be 6 instead of 7. The permutations will be P(6,6) = 6! ways.
But the two vowels A and U can be arranged in two ways, i.e. AU and UA. The required number of arrangements = 2!.6! = 1440 ways.
We may have (1 boy and 3 girls) or (2 boys and 2 girls) or (3 boys and 1 girl) or (4 boys).
Required number of ways =
= = 209.
Required number of 5 digit numbers can be formed by using the digits 1, 0, 2, 3, 5, 6 which are between 50000 and 60000 without repeating the digits are
5 x 4 x 3 x 2 x 1 = 120.
The following arrangements satisfy all 3 conditions.
Arrangement 1: 3 books in a row; 12 rows.
Arrangement 2: 4 books in a row; 9 rows.
Arrangement 3: 6 books in a row; 6 rows.
Arrangement 4: 9 books in a row; 4 rows.
Arrangement 5: 12 books in a row; 3 rows.
Therefore, the possible arrangements are 5.
we can select the 5 member team out of the 8 in 8C5 ways = 56 ways.
The captain can be selected from amongst the remaining 3 players in 3 ways.
Therefore, total ways the selection of 5 players and a captain can be made = 56x3 = 168 ways.
(or)
Alternatively, A team of 6 members has to be selected from the 8 players. This can be done in 8C6 or 28 ways.
Now, the captain can be selected from these 6 players in 6 ways.
Therefore, total ways the selection can be made is 28x6 = 168 ways.
Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4) =( ) = 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! = 120.
Required number of words = (210 x 120) = 25200.
There are 8 students and the maximum capacity of the cars together is 9.
We may divide the 8 students as follows
Case I: 5 students in the first car and 3 in the second
Case II: 4 students in the first car and 4 in the second
Hence, in Case I: 8 students are divided into groups of 5 and 3 in8C3 ways.
Similarly, in Case II: 8 students are divided into two groups of 4 and 4 in 8C4ways.
Therefore, the total number of ways in which 8 students can travel is:
=56 + 70= 126
We first count the number of committee in which
(i). Mr. Y is a member
(ii). the ones in which he is not
Case (i): As Mr. Y agrees to be in committee only where Mrs. Z is a member.
Now we are left with (6-1) men and (4-2) ladies (Mrs. X is not willing to join).
We can choose 1 more in5+
=7 ways.
Case (ii): If Mr. Y is not a member then we left with (6+4-1) people.
we can select 3 from 9 in
=84 ways.
Thus, total number of ways is 7+84= 91 ways.
The given word HAPPYHOLI has 9 letters
These 9 letters can e arranged in 9! ways.
But here in the given word letters H & P are repeated twice each
Therefore, Number of ways these 9 letters can be arranged is
Given Word is PROMISE.
Number of letters in the word PROMISE = 7
Number of ways 7 letters can be arranged = 7! ways
Number of Vowels in word PROMISE = 3 (O, I, E)
Number of ways the vowels can be arranged that 3 Vowels come together = 5! x 3! ways
Now, the number of ways of arrangements so that three vowels should not come together
= 7! - (5! x 3!) ways = 5040 - 720 = 4320.
The number of ways in which the letters of the word RAILINGS can be arranged such that R & S always come together is
Count R & S as only 1 space or letter so that RS or SR can be arranged => 7! x 2!
But in the word RAILINGS, I repeated for 2 times => 7! x 2!/2! = 7! ways = 5040 ways.
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