Permutation & Combination concepts by Gaurav Sharma - Part (1/5)
Director, Genius Tutorials, Karnal ( Haryana ) & Delhi | MSc (Mathematics)
Let us suppose we want to name somebody with three letters say A,B,C. so what could be the possible names?
ABC, ACB, BCA, BAC, CBA, CAB
Now that was easy to figure out as less number of letters are there. But if we have to find that how many 6 letter names can be made out of English alphabets with certain conditions that would be time consuming. Here comes the need and application of this chapter. Permutations and combinations is a very interesting chapter and gives us food for thought
Fundamental principle of Addition
If there are two jobs such that they can be performed independently in m and n ways, then either of the two jobs can be performed in (m + n) ways.
Fundamental principle of Multiplication
If there are two jobs such that one of them can be completed in m ways, and when it has been completed in any one of these m ways, second job can be completed in n ways; then the two jobs in succession can be completed in m x n ways.
How many words of three distinct letters of English alphabets are there?
How many numbers lying between 5000 and 6000 (both inclusive), can be formed if,
(i) Repetition of digits is not allowed.
(ii) Repetition of digits is allowed.
- If we are to form the numbers without repeating any digit then we can’t take 5000 and 6000 as 0 is repeating. Now if number is between 5000 and 6000, we can say that thousand’s place can take only one digit that is 5. For hundred’s place we are left with only 9 digits, for ten’s place only 8 digits and for unit’s place only 7 digits. So the total number of numbers between 5000 and 6000 in which no digit is repeating will be 9 x 8 x 7 = 504
- Now if the repetition of digits is allowed, we can include all the numbers right from 5000 up to and with 6000. So the total number of numbers will be 1001
Number of ways to arrange r thing out of n, nPr = n!/(n-r)!
Number of ways in which 7 students can be seated on 10 chairs in 10P7.
How many words can be formed using all letters of word GENIUS?
Here since we are using 6 letters....so 6 letters can be arranged in 6 place in 6P6 = 6! Ways
How many words can be formed using all letters of the word ORIENTAL
In how many ways can letters of word GENIUS be arranged so that the word must start with N?
Here since first place is fixed as N....so remaining 5 letters can be arranged in 5 places in 5! Ways
How many of the words that can be formed using all letters of the word ORIENTAL will start with N?
How many of the words that can be formed using all letters of the word ORIENTAL will start with N and end with L?
In how many of the words that can be formed using all letters of the word ORIENTAL will all vowels come together?
Now here as we want those words where all vowels come together. so put all vowels in a box and consider it as a single letter. now we have 5 letters and these 5 letters can be arranged in 5! ways and also 4 vowels in that box can also be arranged in 4! ways so answer should be 5!*4!
In how many of the words that can be formed using all letters of the word ORIENTAL all vowels do not come together?
Total words - all vowels together............so 8! - 5!*4!
In how many ways can letters of word GENIUS be arranged so that no two vowels come together?
Here XGXNXSX......so 3 vowels can be arranged in 4 places in 4p3 ways and consonants themselves can be arranged in 3! ways....hence 4p3*3!
In how many of the words that can be formed using all letters of the word ORIENTAL no two vowels come together
In how many of the words that can be formed using all letters of the word ORIENTAL vowels occupy even places?
Now 4 vowels can be arranged at 4 even places in 4! ways and same way 4 consonants in 4! ways at odd places.....so answer should be 4!*4!
In how many words that can be formed using all letters of the word ORIENTAL vowels and constants come alternatively?
As we did for even...same is for odd places...so total should be 2×4! ×4!
Now suppose you have to choose 3 letters from 5 letters then selecting A, B, C and C, B, A are different cases???
Obviously no. here comes the concept of combinations...whenever we need to select few things we use combinations and when we need to arrange we use permutations
Difference between permutation and combination :
- In a combination only selection is made where as in a permutation not only selection is made but also an arrangement in a definite order is considered.
- In a combination, the ordering of the selected objects is immaterial where as in a permutation the ordering is essential. For example A,B and B,A are same as combination but different as permutations
- Practically to find the permutations of n different items, taken r at a time, we first select r items from n items and then arrange them. So, usually the number of permutation exceed the number of combinations.
- Each combination corresponds to many permutations. For example, the six permutations ABC, ACB, BCA, BAC, CBA & CAB correspond to the same combination ABC.
In how many ways can a cricket eleven be chosen from 15 players if
i) a particular player is always chosen
ii) a particular player is never chosen
OA: 14C10, 14C11
The number of permutations of n things of which p1 are alike of one kind; p2 are alike of second kind; p3 are alike of third kind … pr are alike of rth kind such that p1 + p2 … + pr = n is n!/p1!p2!...pr!
Find the number of ways in which 6 boys and 6 girls can be seated in a row so that
- All the girls sit together and all the boys sit together
- All the girls are never together
OA: a) 2*6!6! b) 12!-7!6!
The number of ways in which the letters of the word ARRANGE be arranged so that
- Two R’s are never together
- Two A’s are together but not two R’s
- Neither two A’s nor two R’s are together.
- [RR], A,A,N,G,E = > 6!/2!
and total = > 7!/2!2!
so required answer = 7!/2!2! – 6!/2! = 900.
- Both A’s together - > [AA],R,R,N,G,E = 6!/2! = 360
Both A’s and both R’s together - > [AA],[RR],N,G,E
= 5! = 120
required answer = 360 – 120 = 240.
- Neither two A’s nor two R’s together - > 900 – 240 = 660
How many chords can be drawn through 21 points on a circle ?
A chord is obtained by joining any two points on a circle. Therefore total number of chords drawn through 21 points is same as number of ways of selecting 2 points out of 21 points. Hence number of chords can be drawn in 21C2 = 210 ways.
There are 10 points in a plane, no three of which are in the same straight line, except 4 point which are colinear. Find the
- Number of straight lines that can be formed
- Number of triangles that can be formed using these
- Number of straight lines that can be formed if all 10 are non collinear = 10C2 = 45
But 4 of them are collinear. So number of straight lines that can be formed using 4 points = 4C2 = 6
Now, 4 points when joined pairwise give only one line.
So, total lines = 45 – 6 + 1 = 40
- Number of triangles that can be formed by joining 10 points = 10C3 = 120
Number of triangles formed by joining the 4 points, taken 3 at a time = 4C3 = 4
So total number of triangles = 120 – 4 = 116
N is an integer such that 10^7 ≤ N ≤ 10^8 and sum of digits of N is 3. How many such numbers are possible?
N is an 8 digit number and its left most digit can be 1, 2 or 3.
- When left most digit is 1: Other two digits can be two 1’s and 5 0’s or one 2 and six 0’s.
Number of ways possible for (Two 1s and five 0s): 7!/5!2! = 21
Number of ways possible for (one 2 and 6 0’s): 7!/6! = 7
Therefore, 21 + 7 = 28 cases possible.
- When leftmost digit is 2:
One of the other digit can be 1 so one 1 and six 0’s
Number of ways possible = 7!/1!6! = 7
7 cases possible here.
- When leftmost digit is 3:
All other digits have to be 0
So only 1 case possible
So total cases = 28 + 7 + 1 = 36
There are six boxes numbered 1,2,3… 6. Each box is to be filled up either with a red or a green ball in such a way that atleast one box contains a green ball and the boxes containing green balls are consecutively numbered. The total number of ways in which this can be done is:
Number of ways of filling 1 green ball: 6
Number of ways of filling 2 green balls: 5
Number of ways of filling 3 green balls: 4
[(1,2,3) (2,3,4) (3,4,5) (4,5,6)]
Number of ways of filling 4 green balls: 3
[(1,2,3,4) (2,3,4,5) (3,4,5,6)]
Number of ways of filling 5 green balls: 2
Number of ways of filling 6 green balls : 1
Hence required number of ways = 6 + 5 + 4 + 3 + 2 + 1 = 21
How many words can be formed by taking 4 letters at a time out of the letters of the word MATHEMATICS?
Total Letters : 11 (MM,AA,TT,H,E,I,C,S)
- All distinct : 8C4 x 4! = 1680 ( 8C4 because we have to select 4 out of 8 distinct letters M,A,T,H,E,I,C,S and 4! For the number of ways to arrange these)
- Two distinct and Two alike : 3C1 7C2 x 4!/2! = 756 ( 3C1 to select any 1 pair from (MM,AA,TT) 7C2 to select any two letters from the remaining 7 distinct letters and 4!/2! To arrange them)
- Two alike of one kind and two alike of another kind : 3C2 x 4!/2!2! = 18
So total number of ways = 1680 + 756 + 18 = 2454.
Recommended read: Permutation & Combination concepts by Gaurav Sharma - Part 2