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.
Since each number to be divisible by 5, we must have 5 0r 0 at the units place. But in given digits we have only 5.
So, there is one way of doing it.
Tens place can be filled by any of the remaining 5 numbers.So, there are 5 ways of filling the tens place.
The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.
Required number of numbers = (1 x 5 x 4) = 20.
Let S be the sample space. Then,
n(s) = number of ways of drawing 4 pearls out of 14
=
ways =
= 1001
Let E be the event of drawing 4 pearls of the same colour.
Then, E = event of drawing (4 pearls out of 5) or (4 pearls out of 4) or (4 pearls out of 5)
= 5+1+5 =11
P(E) =
Required probability =
The number of letters in the given word CREATIVITY = 10
Here T & I letters are repeated
=> Number of Words that can be formed from CREATIVITY = 10!/2!x2! = 3628800/4 = 907200
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.
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.
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.
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.
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
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