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.
No. of letters in the word = 6
No. of 'E' repeated = 2
Total No. of arrangement = 6!/2! = 360
The Word LOGARITHMS contains 10 letters.
To find how many 4 letter words we can form from that = =10x9x8x7 = 5040.
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
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
n(E) = 5C3 + 4C3 + 3C3 = 10 + 4 + 1 = 15n(S) = 12C3 = 220
They have to be arranged in the following way :
L T L T L T L T L
The 5 lions should be arranged in the 5 places marked ?L?.
This can be done in 5! ways.
The 4 tigers should be in the 4 places marked ?T?.
This can be done in 4! ways.
Therefore, the lions and the tigers can be arranged in 5!´ 4! ways = 2880 ways.
We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only).
Required number of ways= = 756.
We may have(1 black and 2 non-black) or (2 black and 1 non-black) or (3 black).
Required number of ways= = (45 + 18 + 1) =64
The solution to this problem involves calculating a factorial. Since we want to know how 13 cards can be arranged, we need to compute the value for 13 factorial.
13! = (1)(2)(3)(4)(5)(6)(7)(8)(9)(10)(11)(12)(13) = 6,227,020,800
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