# 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.

28 27501
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

16 27308
Q:

How many lines can you draw using 3 non collinear (not in a single line) points A, B and C on a plane?

 A) 3 B) 6 C) 2 D) 4

Explanation:

You need two points to draw a line. The order is not important. Line AB is the same as line BA. The problem is to select 2 points out of 3 to draw different lines. If we proceed as we did with permutations, we get the following pairs of points to draw lines.

AB , AC

BA , BC

CA , CB

There is a problem: line AB is the same as line BA, same for lines AC and CA and BC and CB.

The lines are: AB, BC and AC ; 3 lines only.

So in fact we can draw 3 lines and not 6 and that's because in this problem the order of the points A, B and C is not important.

58 25492
Q:

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

 A) 53400 B) 17610 C) 11760 D) 45000

Explanation:

Required number of ways = $8C5*10C6$ =  $8C3*10C4$ = 11760

25 25039
Q:

If the letters of the word CHASM are rearranged to form 5 letter words such that none of the word repeat and the results arranged in ascending order as in a dictionary what is the rank of the word CHASM ?

 A) 32 B) 24 C) 72 D) 36

Explanation:

The 5 letter word can be rearranged in 5!=120 Ways without any of the letters repeating.

Then the 25th word will start will CA _ _ _.
The remaining 3 letters can be rearranged in 3!=6 Ways. i.e. 6 words exist that start with CA.

The next word starts with CH and then A, i.e., CHA _ _.
The first of the words will be CHAMS. The next word will be CHASM.

Therefore, the rank of CHASM will be 24+6+2= 32

17 24700
Q:

Find the value of 'n' for which the nth term of two AP'S:

15,12,9.... and -15,-13,-11...... are equal?

 A) n = 2 B) n = 5 C) n = 29/5 D) n = 1

Explanation:

Given are the two AP'S:

15,12,9.... in which a=15, d=-3.............(1)

-15,-13,-11..... in which a'=-15 ,d'=2.....(2)

now using the nth term's formula,we get

a+(n-1)d = a'+(n-1)d'

substituting the value obtained in eq. 1 and 2,

15+(n-1) x (-3) = -15+(n-1) x 2

=> 15 - 3n + 3 = -15 + 2n - 2

=> 12 - 3n = -17 + 2n

=> 12+17 = 2n+3n

=> 29=5n

=> n= 29/5

16 24442
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

22 24385
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.