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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 11
Q11.  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 :  756


We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only).
∴ Required number of ways = ( 7C3 × 6C2 ) + ( 7C4 × 6C1 ) + ( 7C5 )

= ( 7×6×5× 6×5) + ( 7C3 × 6C1 ) + ( 7C2 )
3×2×12×1
= 525 + (7×6×5×6 ) + (7×6 )
3×2×12×1

= ( 525 + 210 + 21 ) = 756.
⚔ DISCUSS

Question 12
Q12.  In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?
 

Answer :  209


We may have (1 boy and 3 girls) or (2 boys and 2 girls) or (3 boys and 1 girl) or (4 boys).
∴ Required number of ways = ( 6C1 × 4C3 ) + ( 6C2 × 4C2 ) + ( 6C3 × 4C1 ) + ( 6C4)
= ( 6C1 × 4C1 ) + ( 6C2 × 4C2 ) + ( 6C3 × 4C1 ) + ( 6C2)

= ( 6 × 4 ) + ( 6×5×4×3 ) + (6×5×4× 4 ) + ( 6×5 )
2×12×13×2×12×1

= ( 24 + 90 + 80 + 15 ) = 209.
⚔ DISCUSS

Question 13
Q13.  A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the box, if at least one black ball is to be included in the draw?
 

Answer :  64


We may have(1 black and 2 non-black) or (2 black and 1 non-black) or (3 black).
∴ Required number of ways = ( 3C1×6C2 ) + ( 3C2×6C1 ) + ( 3C3)

= ( 3 × 6×5 ) + ( 3×2 × 6 ) + 1
2×1 2×1

= ( 45 + 18 + 1 ) = 64.
⚔ DISCUSS

Question 14
Q14.  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 :  20


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 × 5 × 4) = 20.
⚔ DISCUSS

Question 15
Q15.  In how many different ways can the letters of the word 'MATHEMATICS' be arranged so that the vowels always come together?
 

Answer :  120960


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 × 12) = 120960.
⚔ DISCUSS

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