# 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:

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

9 17824
Q:

In how many ways can the letters of the word EDUCATION be rearranged so that the relative position of the vowels and consonants remain the same as in the word EDUCATION?

 A) 4! x 4! B) 5! x 5! C) 4! x 5! D) 3! x 4!

Explanation:

The word EDUCATION is a 9 letter word, with none of the letters repeating.

The vowels occupy 3rd,5th,7th and 8th position in the word and the remaining 5 positions are occupied by consonants

As the relative position of the vowels and consonants in any arrangement should remain the same as in the word EDUCATION, the vowels can occupy only the afore mentioned 4 places and the consonants can occupy1st,2nd,4th,6th and 9th positions.

The 4 vowels can be arranged in the 3rd,5th,7th and 8th position in 4! Ways.

Similarly, the 5 consonants can be arranged in1st,2nd,4th,6th and 9th position in5! Ways.

Hence, the total number of ways = 4! × 5!

16 17434
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

12 17014
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

10 16899
Q:

How many 7 digit numbers can be formed using the digits 1, 2, 0, 2, 4, 2, 4?

 A) 120 B) 360 C) 240 D) 424

Explanation:

There are 7 digits 1, 2, 0, 2, 4, 2, 4 in which 2 occurs 3 times, 4 occurs 2 times.

Number of 7 digit numbers = $7!3!×2!$ = 420

But out of these 420 numbers, there are some numbers which begin with '0' and they are not 7-digit numbers. The number of such numbers beginning with '0'.

=$6!3!×2!$ = 60

Hence the required number of 7 digits numbers = 420 - 60 = 360

21 16773
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.

16 15888
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.

7 15796
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