Out of 15 fruits, 7 are alike of one kind, 5 are alike of a second kind and 3 are alike of a third kind.
Hence, the required number of ways = [ (7+1) (5+1) (3+1) -1] =191
Using, we get
n ? 10 = 12
or, n = 12 + 10 = 22
Let x, y, z be the number of balls received by the three persons, then
Let
x + y + z =21
u + 5 + v + 5 + w + 5 = 21
u + v + w = 6
Total number of solutions =
Given digits are 0, 4, 2, 6
Required 4 digit number should be greater than 6000.
So, first digit must be 6 only and the remaining three places can be filled by one of all the four digits.
This can be done by
1x4x4x4 = 64
Greater than 6000 means 6000 should not be there.
Hence, 64 - 1 = 63.
We are having with digits 2, 3, 4, 5 & 6 and numbers greater than 4000 are to be formed, no digit is repeated.
The number can be 4 digited but greater than 4000 or 5 digited.
Number of 4 digited numbers greater than 4000 are
4 or 5 or 6 can occupy thousand place => 3 x
= 3 x 24 = 72.
5 digited numbers = = 5! = 120
So the total numbers greater than 4000 = 72 + 120 = 192
The first letter is E and the last one is R.
Therefore, one has to find two more letters from the remaining 11 letters.
Of the 11 letters, there are 2 Ns, 2Es and 2As and one each of the remaining 5 letters.
The second and third positions can either have two different letters or have both the letters to be the same.
Case 1: When the two letters are different. One has to choose two different letters from the 8 available different choices. This can be done in 8 * 7 = 56 ways.
Case 2: When the two letters are same. There are 3 options - the three can be either Ns or Es or As. Therefore, 3 ways.
Total number of possibilities = 56 + 3 = 59
A triangle is formed by joining any three non-collinear points in pairs.
There are 7 non-collinear points
The number of triangles formed = = 35
Possibilities Bowlers Batsmen Number of ways
6 9
1 4 7
2 5 6
3 6 5
= 15 x 36 = 540
= 6 x 84 = 504
= 1 x 126 = 126
Total = 1170
Given are the two AP'S:
15,12,9.... in which a=15, d=-3.............(1)
-15,-13,-11..... in which a'=-15 ,d'=2.....(2)
now using the nth term's formula,we get
a+(n-1)d = a'+(n-1)d'
substituting the value obtained in eq. 1 and 2,
15+(n-1) x (-3) = -15+(n-1) x 2
=> 15 - 3n + 3 = -15 + 2n - 2
=> 12 - 3n = -17 + 2n
=> 12+17 = 2n+3n
=> 29=5n
=> n= 29/5
Let 4 girls be one unit and now there are 6 units in all.
They can be arranged in 6! ways.
In each of these arrangements 4 girls can be arranged in 4! ways.
Total number of arrangements in which girls are always together = 6! x 4!= 720 x 24 = 17280
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