All the boxes contain distinct number of chocolates.
For each combination of 4 out of 8 boxes, the box with the greatest number has to be given to the first person, the box with the second highest to the second person and so on.
The number of ways of giving 4 boxes to the 4 person is: 8 = 70
The question requires you to find number of the outcomes in which at most 3 coins turn up as heads.
i.e., 0 coins turn heads or 1 coin turns head or 2 coins turn heads or 3 coins turn heads.
The number of outcomes in which 0 coins turn heads is =1
The number of outcomes in which 1 coin turns head is = =6
The number of outcomes in which 2 coins turn heads is =15
The number of outcomes in which 3 coins turn heads is =20
Therefore, total number of outcomes =1+6+15+20= 42 outcomes
There are 15 dots in total,and to make a triangle we need to select any three of those dots.
So, = 455
- 7 = 14
1 million distinct 3 digit initials are needed.
Let the number of required alphabets in the language be ?n?.
Therefore, using ?n? alphabets we can form n * n * n = distinct 3 digit initials.
Note distinct initials is different from initials where the digits are different.
For instance, AAA and BBB are acceptable combinations in the case of distinct initials while they are not permitted when the digits of the initials need to be different.
This different initials = 1 million
i.e. (1 million = )
=> n = = 100
Hence, the language needs to have a minimum of 100 alphabets to achieve the objective.
There are 4 books on fairy tales and they have to be put together. They can be arranged in 4! ways.
Similarly, there are 5 novels.They can be arranged in 5! ways.
And there are 3 plays.They can be arranged in 3! ways.
So, by the counting principle all of them together can be arranged in 4!´5!´3! ways = 17280
The number of words beign with A is 4!
The number of words beign with E is 4!
The number of words beign with M is 4!
The number of words beign with R is 4!
Number of words beign with VA is 3!
Words beign with VE are VEAMR
VEARM
VEMAR
VEMRA
VERAM
VERMA
Therefore, The Rank of the word VERMA = 4 x 4! + 3! + 6 = 96 + 6 + 6 =108
The Word LOGARITHMS contains 10 letters.
To find how many 4 letter words we can form from that = =10x9x8x7 = 5040.
No. of letters in the word = 6
No. of 'E' repeated = 2
Total No. of arrangement = 6!/2! = 360
In circumcstances word there are 3C's, 2S's, I, U,R, T, A, N, E
Total = 13 letters
But last letter must be N
Hence, available places = 12
In that odd places = 1, 3, 5, 7, 9, 11
Owvels = 4
This can be done in 6P4 ways
Remaining 7 letters can be arranged in 7!/3! x 2! ways
Hence, total number of ways = 6P4 x 7!/3! x 2! = 360 x 5040/12 = 1,51,200 ways.
n(E) = 5C3 + 4C3 + 3C3 = 10 + 4 + 1 = 15n(S) = 12C3 = 220
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