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 a, b, c by taking two at a time are (ab, ba, ac, ca, bc, cb).
   2. All permutations made with the letters a, b, c taking all at a time are:
      ( abc, acb, bac, bca, cab, cba)
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 p_r are alike of r^th kind,
   such that (p₁ + p₂ + ... p_r) = 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 a, b, c taking ab, bc, ca.
   3. The only combination that can be formed of three letters a, b, c 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. ⁿC_n = 1 and ⁿC₀ = 1.
   2. ⁿC_r = 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 6	x	10 x 9 x 8 x 7
	3 x 2 x 1		4 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 5 x 6 x 5 + (7C3 x 6C1) + (7C2) 3 x 2 x 1 2 x 1
	= 525 + 7 x 6 x 5 x 6 + 7 x 6 3 x 2 x 1 2 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.

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