# 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) 25200 | B) 52000 |

C) 120 | D) 24400 |

Explanation:

Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4) = ($7{C}_{3}$*$4{C}_{2}$)

= 210.

Number of groups, each having 3 consonants and 2 vowels = 210.

Each group contains 5 letters.

Number of ways of arranging 5 letters among themselves = 5! = 120

Required number of ways = (210 x 120) = 25200.

A) 720 | B) 520 |

C) 700 | D) 750 |

Explanation:

The word 'LEADING' has 7 different letters.

When the vowels EAI are always together, they can be supposed to form one letter.

Then, we have to arrange the letters LNDG (EAI).

Now, 5 (4 + 1) letters can be arranged in 5! = 120 ways.

The vowels (EAI) can be arranged among themselves in 3! = 6 ways.

Required number of ways = (120 x 6) = 720.

A) 196 | B) 186 |

C) 190 | D) 200 |

Explanation:

When at least 2 women are included.

The committee may consist of 3 women, 2 men : It can be done in $4{C}_{3}*6{C}_{2}$ ways

or, 4 women, 1 man : It can be done in $4{C}_{4}*6{C}_{1}$ways

or, 2 women, 3 men : It can be done in $4{C}_{2}*6{C}_{3}$ ways.

Total number of ways of forming the committees

= $4{C}_{2}*6{C}_{3}+4{C}_{3}*6{C}_{2}+4{C}_{4}*6{C}_{1}$

= 6 x 20 + 4 x 15 + 1x 6

= 120 + 60 + 6 =186

A) 216 | B) 45360 |

C) 1260 | D) 43200 |

Explanation:

There are total 9 letters in the word COMMITTEE in which there are 2M's, 2T's, 2E's.

The number of ways in which 9 letters can be arranged = $\frac{9!}{2!\times 2!\times 2!}$ = 45360

There are 4 vowels O,I,E,E in the given word. If the four vowels always come together, taking them as one letter we have to arrange 5 + 1 = 6 letters which include 2Ms and 2Ts and this be done in $\frac{6!}{2!\times 2!}$ = 180 ways.

In which of 180 ways, the 4 vowels O,I,E,E remaining together can be arranged in $\frac{4!}{2!}$ = 12 ways.

The number of ways in which the four vowels always come together = 180 x 12 = 2160.

Hence, the required number of ways in which the four vowels do not come together = 45360 - 2160 = 43200

A) 360 | B) 700 |

C) 720 | D) 120 |

Explanation:

The word 'OPTICAL' contains 7 different letters.

When the vowels OIA are always together, they can be supposed to form one letter.

Then, we have to arrange the letters PTCL (OIA).

Now, 5 letters can be arranged in 5! = 120 ways.

The vowels (OIA) can be arranged among themselves in 3! = 6 ways.

Required number of ways = (120 x 6) = 720.

A) 1260 | B) 1400 |

C) 1250 | D) 1600 |

Explanation:

A team of 6 members has to be selected from the 10 players. This can be done in $10{C}_{6}$ or 210 ways.

Now, the captain can be selected from these 6 players in 6 ways.

Therefore, total ways the selection can be made is 210×6= 1260

A) 135 | B) 63 |

C) 125 | D) 64 |

Explanation:

Required number of ways = $(7{C}_{5}*3{C}_{2})=(7{C}_{2}*3{C}_{1})=63$

A) 601 | B) 600 |

C) 603 | D) 602 |

Explanation:

If the word started with the letter A then the remaining 5 positions can be filled in 5! Ways.

If it started with c then the remaining 5 positions can be filled in 5! Ways.Similarly if it started with H,I,N the remaining 5 positions can be filled in 5! Ways.

If it started with S then the remaining position can be filled with A,C,H,I,N in alphabetical order as on dictionary.

The required word SACHIN can be obtained after the 5X5!=600 Ways i.e. SACHIN is the 601th letter.