permutation & combination

Permutation and Combination

• Permutation : It is the different arrangements of a given number of elements taken one by one, or some, or all at a time. For example, if we have two elements A and B, then there are two possible arrangements, AB and BA.
• Number of permutations when ‘r’ elements are arranged out of a total of ‘n’ elements is n Pr = n! / (n – r)!. For example, let n = 4 (A, B, C and D) and r = 2 (All permutations of size 2). The answer is 4!/(4-2)! = 12. The twelve permutations are AB, AC, AD, BA, BC, BD, CA, CB, CD, DA, DB and DC.
• Combination : It is the different selections of a given number of elements taken one by one, or some, or all at a time. For example, if we have two elements A and B, then there is only one way select two items, we select both of them.
• Number of combinations when ‘r’ elements are selected out of a total of ‘n’ elements is n C r = n! / [ (r !) x (n – r)! ]. For example, let n = 4 (A, B, C and D) and r = 2 (All combinations of size 2). The answer is 4!/((4-2)!*2!) = 6. The six combinations are AB, AC, AD, BC, BD, CD.
• n C r = n C (n – r)
NOTE : In the same example, we have different cases for permutation and combination. For permutation, AB and BA are two different things but for selection, AB and BA are same.

Sample Problems

Question 1 : How many words can be formed by using 3 letters from the word “DELHI” ?
Solution : The word “DELHI” has 5 different words.
Therefore, required number of words = 5 P 3 = 5! / (5 – 3)!
=> Required number of words = 5! / 2! = 120 / 2 = 60

Question 2 : How many words can be formed by using the letters from the word “DRIVER” such that all the vowels are always together ?
Solution : In these type of questions, we assume all the vowels to be a single character, i.e., “IE” is a single character.
So, now we have a total of 5 characters in the word, namely, D, R, V, R, IE.
But, R occurs 2 times.
=> Number of possible arrangements = 5! / 2! = 60
Now, the two vowels can be arranged in 2! = 2 ways.
=> Total number of possible words such that the vowels are always together= 60 x 2 = 120

Question 3 : In how many ways, can we select a team of 4 students from a given choice of 15 ?
Solution : Number of possible ways of selection = 15 C 4 = 15 ! / [(4 !) x (11 !)]
=> Number of possible ways of selection = (15 x 14 x 13 x 12) / (4 x 3 x 2 x 1) = 1365

Question 4 : In how many ways can a group of 5 members be formed by selecting 3 boys out of 6 and 2 girls out of 5 ?
Solution : Number of ways 3 boys can be selected out of 6 = 6 C 3 = 6 ! / [(3 !) x (3 !)] = (6 x 5 x 4) / (3 x 2 x 1) = 20
Number of ways 2 girls can be selected out of 5 = 5 C 2 = 5 ! / [(2 !) x (3 !)] = (5 x 4) / (2 x 1) = 10
Therefore, total number of ways of forming the group = 20 x 10 = 200

Question 5 : How many words can be formed by using the letters from the word “DRIVER” such that all the vowels are never together ?
Solution : we assume all the vowels to be a single character, i.e., “IE” is a single character.
So, now we have a total of 5 characters in the word, namely, D, R, V, R, IE.
But, R occurs 2 times.
=> Number of possible arrangements = 5! / 2! = 60
Now, the two vowels can be arranged in 2! = 2 ways.
=> Total number of possible words such that the vowels are always together = 60 x 2 = 120
Also, total number of possible words = 6! / 2! = 720 / 2 = 360
Therefore, total number of possible words such that the vowels are never together = 360 – 120 = 240
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Sometimes students find it difficult to decide whether a problem is on permutation or combination or both. Based on certain words/phrases occurring in the problem we can fairly decide its nature as per the following table :
PROBLEMS  OF COMBINATIONS 
•Selections , choose
•Distributed group is formed
•committee
•Geometrical problems
PROBLEMS  OF PERMUTATIONS

•Arrangements
•Standing in a line seated in a row
•problems on digits


•Problems on letters from a word
once you complete NCERT the important topics for you to study
Circular permutations
Combination:


EXPONENT OF PRIME P IN n! & problems based on GEOMETRY


Combination for jee main & advanced


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