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⁞ Permutations And Combinations

Aptitude ➤ Permutations And Combinations ➤ 1

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:

i. We define 0! = 1.
ii. 4! = (4 x 3 x 2 x 1) = 24.
iii. 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:

i. All permutations (or arrangements) made with the letters a, b, c by taking two at a time are (ab, ba, ac, ca, bc, cb).
ii. 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:

i. 6P2 = (6 x 5) = 30.
ii. 7P3 = (7 x 6 x 5) = 210.
iii. 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:
i. 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.
ii. All the combinations formed by a, b, c taking ab, bc, ca.
iii. The only combination that can be formed of three letters a, b, c taken all at a time is abc.
iv. Various groups of 2 out of four persons A, B, C, D are:
AB, AC, AD, BC, BD, CD.
v. 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:

i. nCn = 1 and nC0 = 1.
ii. 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

Question 6
Q6.  How many words can be formed from the letters of the word 'EXTRA' so that the vowels are never together?
 

Answer :  72


The given word contains 5 different letters.
Taking the vowels EA together, we treat them as one letter.
Then, the letters to be arranged are XTR (EA).
These letters can be arranged in 4! = 24 ways.
The vowels EA may be arranged amongst themselves in 2! = 2 ways.

Number of words, each having vowels together = (24x2) = 48 ways.

Total number of words formed by using all the letters of the given words = 5! = (5x4x3x2x1) = 120.

∴ Number of words, each having vowels never together = (120-48) = 72.
⚔ DISCUSS

Question 7
Q7.  In how many different ways can the letters of the word 'DETAIL' be arranged in such a way that the vowels occupy only the odd positions?
 

Answer :  36


There are 6 letters in the given word, out of which there are 3 vowels and 3 consonants.
Let us mark these positions as under:

(1) (2) (3) (4) (5) (6)

Now, 3 vowels can be placed at any of the three places out 4, marked 1, 3, 5.
Number of ways of arranging the vowels = 3P3 = 3! = 6.
Also, the 3 consonants can be arranged at the remaining 3 positions.
Number of ways of these arrangements = 3P3 = 3! = 6.

Total number of ways = (6 x 6) = 36.
⚔ DISCUSS

Question 8
Q8.  How many words can be formed from the letters of the word 'DIRECTOR' so that the vowels are always together?
 

Answer :  2160


In the given word, we treat the vowels IEO as one letter.
Thus, we have DRCTR (IEO).
This group has 6 letters of which R occurs 2 times and others are different.

Number of ways arranging these letters =6!= 360.
2!

Now 3 vowels can be arranged among themselves in 3! = 6 ways.

∴ Required number of ways = (360x6) = 2160.
⚔ DISCUSS

Question 9
Q9.  In how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women?
 

Answer :  63


Required number of ways = (7C5 x 3C2)

=7!x3!
5!(7-5)! 2!(3-2)!
= 7!x3!
5!(2!) 2!(1!)
=7x6 x 3 = 63.
2x1


⚔ DISCUSS

Question 10
Q10.  In how many ways a committee, consisting of 5 men and 6 women can be formed from 8 men and 10 women?
 

Answer :  11760


Required number of ways = ( 8C5 × 10C6 )
= ( 8C3 × 10C4 )

= (8×7×6×10×9×8×7) = 11760
3×2×14×3×2×1


⚔ DISCUSS

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