# 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) $P_{2}^{6}=\left(6\times 5\right)=30$ (ii) $P_{3}^{7}=\left(7\times 6\times 5\right)=210$

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 :

$\frac{n!}{({p}_{1}!)\times ({p}_{2}!)....({p}_{r}!)}$

**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:

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

Note : (i)$C_{n}^{n}=1andC_{0}^{n}=1$ (ii)$C_{r}^{n}=C_{(n-r)}^{n}$

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

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!

A) 36 | B) 25 |

C) 42 | D) 120 |

Explanation:

There are 6 letters in the given word, out of which there are 3 vowels and 3 consonants.

Let us mark these positions as under:

(1) (2) (3) (4) (5) (6)

Now, 3 vowels can be placed at any of the three places out 4, marked 1, 3, 5.

Number of ways of arranging the vowels = $3{P}_{3}$ = 3! = 6.

Also, the 3 consonants can be arranged at the remaining 3 positions.

Number of ways of these arrangements = $3{P}_{3}$ = 3! = 6.

Total number of ways = (6 x 6) = 36.

A) 53400 | B) 17610 |

C) 11760 | D) 45000 |

Explanation:

Required number of ways = $\left(8{C}_{5}*10{C}_{6}\right)$ = $\left(8{C}_{3}*10{C}_{4}\right)$ = 11760

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 $4{P}_{3}$ ways and 4 constants can be arranged in $4{P}_{4}$ ways.

Number of words =$4{P}_{3}$ x $4{P}_{4}$= 24 x 24 = 576

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.

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 = $\frac{7!}{3!\times 2!}$ = 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'.

=$\frac{6!}{3!\times 2!}$ = 60

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

A) 525 | B) 535 |

C) 545 | D) 555 |

Explanation:

The number of points of intersection of 37 lines is $C_{2}^{37}$. But 13 straight lines out of the given 37 straight lines pass through the same point A.

Therefore instead of getting $C_{2}^{13}$ 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 $C_{2}^{11}$ points, we get only one point B.

Hence the number of intersection points of the lines is $C_{2}^{37}-C_{2}^{13}-C_{2}^{11}+2$ = 535

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.