PERMUTATIONS & COMBINATIONS

Permutations and Combinations



  1. Factorial Notation:
    Let n be a positive integer. Then, factorial n, denoted n! is defined as:
    n! = n(n - 1)(n - 2) ... 3.2.1.
    Examples:
    1. We define 0! = 1.
    2. 4! = (4 x 3 x 2 x 1) = 24.
    3. 5! = (5 x 4 x 3 x 2 x 1) = 120.
  2. Permutations:
    The different arrangements of a given number of things by taking some or all at a time, are called permutations.
    Examples:
    1. All permutations (or arrangements) made with the letters abc by taking two at a time are (abbaaccabccb).
    2. All permutations made with the letters abc taking all at a time are:
      ( abcacbbacbcacabcba)
  3. Number of Permutations:
    Number of all permutations of n things, taken r at a time, is given by:
    nPr = n(n - 1)(n - 2) ... (n - r + 1) =n!
    (n - r)!
    Examples:
    1. 6P2 = (6 x 5) = 30.
    2. 7P3 = (7 x 6 x 5) = 210.
    3. Cor. number of all permutations of n things, taken all at a time = n!.
  4. An Important Result:
    If there are n subjects of which p1 are alike of one kind; p2 are alike of another kind;p3 are alike of third kind and so on and pr are alike of rth kind,
    such that (p1 + p2 + ... pr) = n.
    Then, number of permutations of these n objects is =n!
    (p1!).(p2)!.....(pr!)
  5. Combinations:
    Each of the different groups or selections which can be formed by taking some or all of a number of objects is called a combination.
    Examples:
    1. Suppose we want to select two out of three boys A, B, C. Then, possible selections are AB, BC and CA.
      Note: AB and BA represent the same selection.
    2. All the combinations formed by abc taking abbcca.
    3. The only combination that can be formed of three letters abc taken all at a time is abc.
    4. Various groups of 2 out of four persons A, B, C, D are:
      AB, AC, AD, BC, BD, CD.
    5. Note that ab ba are two different permutations but they represent the same combination.
  6. Number of Combinations:
    The number of all combinations of n things, taken r at a time is:
    nCr =n!=n(n - 1)(n - 2) ... to r factors.
    (r!)(n - r)!r!
    Note:
    1. nCn = 1 and nC0 = 1.
    2. nCr = nC(n - r)
    Examples:
    i.   11C4 =(11 x 10 x 9 x 8)= 330.
    (4 x 3 x 2 x 1)
    ii.   16C13 = 16C(16 - 13) = 16C3 =16 x 15 x 14=16 x 15 x 14= 560.
    3!3 x 2 x 1

Example:

In how many ways a committee, consisting of 5 men and 6 women can be formed from 8 men and 10 women?

Answer:
Required number of ways= (8C5 x 10C6)= (8C3 x 10C4)
8 x 7 x 6x10 x 9 x 8 x 7
3 x 2 x 14 x 3 x 2 x 1
= 11760.

Example:

In how many ways can the letters of the word 'LEADER' be arranged?

Answer:

The word 'LEADER' contains 6 letters, namely 1L, 2E, 1A, 1D and 1R.
 Required number of ways =6!= 360.
(1!)(2!)(1!)(1!)(1!)

Example:

From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?

Answer:

We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only).
 Required number of ways= (7C3 x 6C2) + (7C4 x 6C1) + (7C5)
=7 x 6 x 5x6 x 5+ (7C3 x 6C1) + (7C2)
3 x 2 x 12 x 1
= 525 +7 x 6 x 5x 6+7 x 6
3 x 2 x 12 x 1
= (525 + 210 + 21)
= 756.

Example:

How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated?

Answer:

Since each desired number is divisible by 5, so we must have 5 at the unit place. So, there is 1 way of doing it.
The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place.
The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.
 Required number of numbers = (1 x 5 x 4) = 20.

Example:

In how many different ways can the letters of the word 'MATHEMATICS' be arranged so that the vowels always come together?

Answer:

In the word 'MATHEMATICS', we treat the vowels AEAI as one letter.
Thus, we have MTHMTCS (AEAI).
Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice and the rest are different.
 Number of ways of arranging these letters =8!= 10080.
(2!)(2!)
Now, AEAI has 4 letters in which A occurs 2 times and the rest are different.
Number of ways of arranging these letters =4!= 12.
2!
 Required number of words = (10080 x 12) = 120960.

sujan koilakuntla

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