# 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:

In a Plane there are 37 straight lines, of which 13 pass through the point A and 11 pass through the point B. Besides, no three lines pass through one point, no lines passes through both points A and B , and no two are parallel. Find the number of points of intersection of the straight lines.

 A) 525 B) 535 C) 545 D) 555

Explanation:

The number of points of intersection of 37 lines is $C237$. But 13 straight lines out of the given 37 straight lines pass through the same point A.

Therefore instead of getting $C213$ points, we get only one point A. Similarly 11 straight lines out of the given 37 straight lines intersect at point B. Therefore instead of getting $C211$ points, we get only one point B.

Hence the number of intersection points of the lines is  = 535

41 22953
Q:

How many ways can 10 letters be posted in 5 post boxes, if each of the post boxes can take more than 10 letters ?

 A) 5^10 B) 10^5 C) 5P5 D) 5C5

Explanation:

Each of the 10 letters can be posted in any of the 5 boxes.

So, the first letter has 5 options, so does the second letter and so on and so forth for all of the 10 letters.

i.e. 5*5*5*….*5 (upto 10 times) = 5 ^ 10.

14 22101
Q:

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?

 A) 209 B) 290 C) 200 D) 208

Explanation:

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 = $6C1*4C3+6C2*4C2+6C3*4C1+6C4$

$6C1*4C1+6C2*4C2+6C3*4C1+6C2$ = 209.

11 21946
Q:

How many arrangements of the letters of the word ‘BENGALI’ can be made if the vowels are to occupy only odd places.

 A) 720 B) 576 C) 567 D) 625

Explanation:

There are 7 letters in the word Bengali of these 3 are vowels and 4 consonants.

There are 4 odd places and 3 even places. 3 vowels can occupy 4 odd places in $4P3$ ways and 4 constants can be arranged in $4P4$ ways.

Number of words =$4P3$  x $4P4$= 24 x 24 = 576

17 21399
Q:

There are 7 non-collinear points. How many triangles can be drawn by joining these points?

 A) 10 B) 30 C) 35 D) 60

Explanation:

A triangle is formed by joining any three non-collinear points in pairs.

There are 7 non-collinear points

The number of triangles formed = $7C3$ = 35

15 21155
Q:

How many different four letter words can be formed (the words need not be meaningful using the letters of the word "MEDITERRANEAN" such that the first letter is E and the last letter is R?

 A) 59 B) 56 C) 64 D) 55

Explanation:

The first letter is E and the last one is R.

Therefore, one has to find two more letters from the remaining 11 letters.

Of the 11 letters, there are 2 Ns, 2Es and 2As and one each of the remaining 5 letters.

The second and third positions can either have two different letters or have both the letters to be the same.

Case 1: When the two letters are different. One has to choose two different letters from the 8 available different choices. This can be done in 8 * 7 = 56 ways.

Case 2: When the two letters are same. There are 3 options - the three can be either Ns or Es or As. Therefore, 3 ways.

Total number of possibilities = 56 + 3 = 59

21 20883
Q:

A polygon has 44 diagonals, then the number of its sides are ?

 A) 13 B) 9 C) 11 D) 7

Explanation:

Let the number of sides be n.

The number of diagonals is given by nC2 - n

Therefore, nC2 - n = 44, n>0

nC2 - n = 44

n- 3n - 88 = 0

n2 -11n + 8n - 88 = 0

n(n - 11) + 8(n - 11) = 0

n = -8 or n = 11.

As n>0, n will not be -8. Therefore, n=11.

24 20332
Q:

5 men and 4 women are to be seated in a row so that the women occupy the even places . How many such arrangements are possible?

 A) 2880 B) 1440 C) 720 D) 2020

Explanation:

There are total 9 places out of which 4 are even and rest 5 places are odd.

4 women can be arranged at 4 even places in 4! ways.

and 5 men can be placed in remaining 5 places in 5! ways.

Hence, the required number of permutations  = 4! x 5! = 24 x 120 = 2880