Placements | QA | Permutation and Combination

2.3
  • Factorial : The factorial of ‘n’, written as n ! = n x (n-1) x (n-2) x (n-3) x … x 3 x 2 x 1
    0 ! = 1
  • 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 the possible arrangements are AB and BA. So, total 2 arrangements are possible.
  • Number of permutations when ‘r’ elements are arranged out of a total of ‘n’ elements is n P r = n! / (n – r)!
  • 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 the possible selections are AB or BA. So, total 1 arrangements are possible.
  • Number of combinations when ‘r’ elements are selected out of a total of ‘n’ elements is n C r = n! / [ (r !) x (n – r)! ]
  • 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
     
    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.
     
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