We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only).
Required number of ways= = 756.
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.
n(E) = 5C3 + 4C3 + 3C3 = 10 + 4 + 1 = 15n(S) = 12C3 = 220
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.
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
Since there are 5 cups of each kind,prepared with milk or tea leaves added first,are identical hence,total number of different people ways of presenting the cups to the expert is 10!/(5! x 5!)= 252
The remainder on the first card can be 0,1,2 or 3 i.e 4 possibilities.
The remainder of the number on the next card when divided by 4 can have 3 possible values (except the one occurred earlier).
For each value on the card the remainder can have 3 possible values.
The total number of possible sequences is: 4*3^4
Let the number of Rose plants be ?a?.
Let number of marigold plants be ?b?.
Let the number of Sunflower plants be ?c?.
20a+5b+1c=1000; a+b+c=100
Solving the above two equations by eliminating c,
19a+4b=900
b = (900-19a)/4
b = 225 - 19a/4----------(1)
b being the number of plants, is a positive integer, and is less than 99, as each of the other two types have at least one plant in the combination i.e .:0 < b < 99--------(2)
Substituting (1) in (2),
0 < 225 - 19a/4 < 99
225 < -19a/4 < (99 -225)
=> 4 x 225 > 19a > 126 x 4
=> 900/19 > a > 505
a is the integer between 47 and 27 ----------(3)
From (1), it is clear, a should be multiple of 4.
Hence possible values of a are (28,32,36,40,44)
For a=28 and 32, a+b>100
For all other values of a, we get the desired solution:
a=36,b=54,c=10
a=40,b=35,c=25
a=44,b=16,c=40
Three solutions are possible.
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