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

    Quiz on Permutation and Combination
    Practice Questions on Permutation and Combination.

    This article has been contributed by Nishant Arora. Please write comments if you have any doubts related to the topic discussed above, or if you are facing difficulty in any question or if you would like to discuss a question other than those mentioned above.
     
    Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above



    My Personal Notes arrow_drop_up

    Article Tags :


    Be the First to upvote.


    Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.