# Permutations and Combinations Questions

FACTS  AND  FORMULAE  FOR  PERMUTATIONS  AND  COMBINATIONS  QUESTIONS

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 : We define 0! = 1.

4! = (4 x 3 x 2 x 1) = 24.

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.

Ex1 : All permutations (or arrangements) made with the letters a, b, c by taking two at a time are (ab, ba, ac, ca, bc, cb).

Ex2 : All permutations made with the letters a, b, c taking all at a time are:( abc, acb, bac, bca, cab, cba)

Number of Permutations: Number of all permutations of n things, taken r at a time, is given by:

$P_{r}^{n}=n\left(n-1\right)\left(n-2\right)....\left(n-r+1\right)=\frac{n!}{\left(n-r\right)!}$

Ex : (i) $P_{2}^{6}=\left(6×5\right)=30$   (ii) $P_{3}^{7}=\left(7×6×5\right)=210$

Cor. number of all permutations of n things, taken all at a time = n!.

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 $\left({p}_{1}+{p}_{2}+...+{p}_{r}\right)=n$

Then, number of permutations of these n objects is :

3.  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.

Ex.1 : Suppose we want to select two out of three boys A, B, C. Then, possible selections are AB, BC and CA.

Note that AB and BA represent the same selection.

Ex.2 : All the combinations formed by a, b, c taking ab, bc, ca.

Ex.3 : The only combination that can be formed of three letters a, b, c taken all at a time is abc.

Ex.4 : Various groups of 2 out of four persons A, B, C, D are : AB, AC, AD, BC, BD, CD.

Ex.5 : Note that ab ba are two different permutations but they represent the same combination.

Number of Combinations: The number of all combinations of n things, taken r at a time is:

Note : (i)     (ii)$C_{r}^{n}=C_{\left(n-r\right)}^{n}$

Examples : (i) $C_{4}^{11}=\frac{11×10×9×8}{4×3×2×1}=330$      (ii)$C_{13}^{16}=C_{\left(16-13\right)}^{16}=C_{3}^{16}=560$

Q:

A letter lock consists of three rings each marked with six different letters. The number of distinct unsuccessful attempts to open the lock is at the most  ?

 A) 215 B) 268 C) 254 D) 216

Explanation:

Since each ring consists of six different letters, the total number of attempts possible with the three rings is = 6 x 6 x 6 = 216. Of these attempts, one of them is a successful attempt.

Maximum number of unsuccessful attempts = 216 - 1 = 215.

14 12758
Q:

From 5 consonants and 4 vowels, how many words can be formed using 3 consonants and 2 vowels ?

 A) 7600 B) 7200 C) 6400 D) 3600

Explanation:

From 5 consonants, 3 consonants can be selected in $5C3$ ways.

From 4 vowels, 2 vowels can be selected in $4C2$ ways.

Now with every selection, number of ways of arranging 5 letters is $5P5$ways.

Total number of words = $5C3*4C2*5P5$

= 10x 6 x 5 x 4 x 3 x 2 x 1= 7200

3 12700
Q:

In how many ways can 5 different toys be packed in 3 identical boxes such that no box is empty, if any of the boxes may hold all of the toys ?

 A) 36 B) 25 C) 24 D) 72

Explanation:

The toys are different; The boxes are identical

If none of the boxes is to remain empty, then we can pack the toys in one of the following ways

a. 2, 2, 1

b. 3, 1, 1

Case a. Number of ways of achieving the first option 2 - 2 - 1

Two toys out of the 5 can be selected in $5C2$ ways. Another 2 out of the remaining 3 can be selected in $3C2$ ways and the last toy can be selected in $1C1$ way.

However, as the boxes are identical, the two different ways of selecting which box holds the first two toys and which one holds the second set of two toys will look the same. Hence, we need to divide the result by 2

Therefore, total number of ways of achieving the 2 - 2 - 1 option is ways $5C2*3C2$= 15 ways

Case b. Number of ways of achieving the second option 3 - 1 - 1

Three toys out of the 5 can be selected in $5C3$ ways. As the boxes are identical, the remaining two toys can go into the two identical looking boxes in only one way.

Therefore, total number of ways of getting the 3 - 1 - 1 option is $5C3$ = 10 = 10 ways.

Total ways in which the 5 toys can be packed in 3 identical boxes

= number of ways of achieving Case a + number of ways of achieving Case b= 15 + 10 = 25 ways.

20 12273
Q:

 A) 18! B) 18! x 19! C) 18!(6 x 24) D) 18! x 30

Explanation:

Here clock-wise and anti-clockwise arrangements are same.

Hence total number of circular–permutations: $18P122*12$ = $18!6*24$

21 12075
Q:

There are 6 bowlers and 9 batsmen in a cricket club. In how many ways can a team of 11 be selected so that the team contains at least 4 bowlers?

 A) 1170 B) 1200 C) 720 D) 360

Explanation:

Possibilities     Bowlers      Batsmen         Number of ways

6               9

1              4                7              $(6C4*9C7)$

2              5                6              $6C5*9C6$

3              6                5              $6C6*9C5$

$6C4*9C7$ = 15 x 36 = 540

$6C5*9C6$ = 6 x 84 = 504

$6C6*9C5$= 1 x 126 = 126

Total = 1170

11 11355
Q:

In how many ways can 4 girls and 5 boys be arranged in a row so that all the four girls are together ?

 A) 18000 B) 17280 C) 17829 D) 18270

Explanation:

Let 4 girls be one unit and now there are 6 units in all.

They can be arranged in 6! ways.

In each of these arrangements 4 girls can be arranged in 4! ways.

Total number of arrangements in which girls are always together = 6! x 4!= 720 x 24 = 17280

3 10622
Q:

A coach must choose five starters from a team of 12 players. How many different ways can the coach choose the starters ?

 A) 569 B) 729 C) 625 D) 769

Explanation:

Choose 5 starters from a team of 12 players. Order is not important.

$12C5$= 729

3 10402
Q:

A team of 8 students goes on an excursion, in two cars, of which one can seat 5 and the other only 4. In how many ways can they travel?

 A) 126 B) 120 C) 146 D) 156

Explanation:

There are 8 students and the maximum capacity of the cars together is 9.

We may divide the 8 students as follows

Case I: 5 students in the first car and 3 in the second
Case II: 4 students in the first car and 4 in the second

Hence, in Case I: 8 students are divided into groups of 5 and 3 in8C3 ways.

Similarly, in Case II: 8 students are divided into two groups of 4 and 4 in 8C4ways.

Therefore, the total number of ways in which 8 students can travel is:
$8C3+8C4$=56 + 70= 126