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
Let the number of sides be n.
The number of diagonals is given by nC2 - n
Therefore, nC2 - n = 44, n>0
nC2 - n = 44
n2 - 3n - 88 = 0
n2 -11n + 8n - 88 = 0
As n>0, n will not be -8. Therefore, n=11.
Required Number of ways = = 36
The word EDUCATION is a 9 letter word, with none of the letters repeating.
The vowels occupy 3rd,5th,7th and 8th position in the word and the remaining 5 positions are occupied by consonants
As the relative position of the vowels and consonants in any arrangement should remain the same as in the word EDUCATION, the vowels can occupy only the afore mentioned 4 places and the consonants can occupy1st,2nd,4th,6th and 9th positions.
The 4 vowels can be arranged in the 3rd,5th,7th and 8th position in 4! Ways.
Similarly, the 5 consonants can be arranged in1st,2nd,4th,6th and 9th position in5! Ways.
Hence, the total number of ways = 4! × 5!
Using, we get
n ? 10 = 12
or, n = 12 + 10 = 22
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
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
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