Given 11 questions of type True or False
Then, Each of these questions can be answered in 2 ways (True or false)
Therefore, no. of ways of answering 11 questions = = 2048 ways.
Given that there are three blue marbles, four red marbles, six green marbles and two yellow marbles.
Probability that both marbles are blue = ³C?/¹?C? = (3 x 2)/(15 x 14) = 1/35
Probability that both are yellow = ²C?/¹?C? = (2 x 1)/(15 x 14) = 1/105
Probability that one blue and other is yellow = (³C? x ²C?)/¹?C? = (2 x 3 x 2)/(15 x 14) = 2/35
Required probability = 1/35 + 1/105 + 2/35 = 3/35 + 1/105 = 1/35(3 + 1/3) = 10/(3 x 35) = 2/21
15! - 14! - 13!
= (15 × 14 × 13!) - (14 × 13!) - (13!)
= 13! (15 × 14 - 14 - 1)
= 13! (15 × 14 - 15)
= 13! x 15 (14 - 1)
= 15 × 13 × 13!
Given letters are k, l, m, n, o = 5
number of letters to be in the words = 3
Total number of words that can be formed from these 5 letters taken 3 at a time without repetation of letters =
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
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 =
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.
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.
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.
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