In how many ways can the letters of the word DELHI be arranged?
A and B can occupy the first and the ninth places, the second and the tenth places, the third and the eleventh place and so on... This can be done in 18 ways.
A and B can be arranged in 2 ways.
All the other 24 alphabets can be arranged in 24! ways.
Hence the required answer = 2 x 18 x 24!
How many numbers between 200 and 1200 can be formed with the digits 0, 1, 2, 3 (repetition of digits not allowed?
So total of 14 numbers (without repetition) between 200 & 1200 with 0, 1, 2 & 3
In how many ways can 10 identical presents be distributed among 6 children so that each child gets at least one present?
We have to count natural numbers which have a maximum of 4 digits. The required answer will be given by: Number of single digit numbers + Number of two digit numbers + Number of three digit numbers + Number of four digit numbers.
A captain and a vice-captain are to be chosen out of a team having eleven players. How many ways are there to achieve this?
Explaination: out of 11 player 1 captain can be choose 11 ways, Now remaining 10 player,wise captain can be choose in 10 ways Therefore total number of ways =11*10=110 ways
In how many ways can Ram choose a vowel and a consonant from the letters of the word ALLAHABAD?
How many motor vehicle registration number of 4 digits can be formed with the digits 0, 1,2, 3, 4, 5? (No digit being repeated.)
Total number of 6 digit numbers having 3 odd and 3 even digits (including zero in the left most place) = 53 x 53.
From this subtract the number of 5 digit numbers with 2 even digits and 3 odd digits (to take care of the extra counting due to zero)
There are ten subjects in the school day at St.Vincent’s High School but the sixth standard students have only 5 periods in a day. In how many ways can we form a time table for the day for the sixth standard students if no subject is repeated?
The possible cases for counting are: Number of numbers when the units digit is nine + the number of numbers when neither the units digit nor the left most is nine + number of numbers when the left most digit is nine.
How many batting orders are possible for the Indian cricket team if there is a squad of 15 to choose from such that Sachin Tendulkar is always chosen?
The condition is that we have to count the number of natural numbers not more than 4300.
The total possible numbers with the given digits = 5x5 x5 x5 = 625 - 1 = 624.
Subtract form this the number of natural number greater than 4300 which can be formed from the given digits =1x 2x 5x5 -1 =49 .
Hence, the required number of numbers = 624 - 49 = 575.
How many distinct words can be formed out of the word PROWLING which start with R & end with W?
How many even numbers of four digits can be formed with the digits 1, 2, 3, 4, 5, 6 (repetitions of digits are allowed)?
The number of numbers formed would be given by 5 x 4 x 3 (given that the first digit can be filled in 5 ways, the second in 4 ways and the third in 3 ways - MNP rule).
In the above question, what will be the number of ways of selecting the committee with at least 3 women such that at least one woman holds the post of either a president or a vice-president?
For a straight line we just need to select 2 points out of the 8 points available. 8C2 would be the number of ways of doing this.
In how many ways can the letters of the word ‘EQUATION’ be arranged so that all the vowels come together?
How many quadrilateral can be formed from 25 points out of which 7 are collinear
otal number of quadrilateral combination possible if none of the points are collinear = 25C4 = 12650
If we form a geometry by joining any three points out of seven collinear points and one point from 18 non collinear points, it will give us a triangle instead of quadrilateral. So we have to eliminate number of combinations which can be formed in this way, which is 7C3 x 18C1 = 35 x 18 = 630
We also can't form quadrilateral if we choose all four vertices of quadrilateral to be any 4 points from 7 collinear points. It will come out to be a straight line. So we have to eliminate such combinations also. Which is 7C4 = 35
So net number of possible quadrilaterals = 12650 - 630 - 35 = 11985
In how many ways a committee consisting of 5 men and 3 women, can be chosen from 9 men and 12 women.
Choose 5 men out of 9 men = 9C5 ways = 126 ways
Choose 3 women out of 12 women = 12C3 ways = 220 ways
The committee can be chosen in 27720 ways