# 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) 17756 | B) 17576 |

C) 12657 | D) 12666 |

Explanation:

The first letter from the right can be chosen in 26 ways because there are 26 alphabets.

Having chosen this, the second letter can be chosen in 26 ways

The first two letters can chosen in 26 x 26 = 676 ways

Having chosen the first two letters, the third letter can be chosen in 26 ways.

All the three letters can be chosen in 676 x 26 =17576 ways.

It implies that the maximum possible number of five letter palindromes is 17576 because the fourth letter is the same as the second letter and the fifth letter is the same as the first letter.

A) 500 | B) 490 |

C) 495 | D) 540 |

Explanation:

For any set of 4 points we get a cyclic quadrilateral. Number of ways of choosing 4 points out of 12 points is $12{C}_{4}$ = 495.

Therefore, we can draw 495 quadrilaterals

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 = $\left(6{C}_{1}*4{C}_{3}\right)+\left(6{C}_{2}*4{C}_{2}\right)+\left(6{C}_{3}*4{C}_{1}\right)+6{C}_{4}$

= $\left(6{C}_{1}*4{C}_{1}\right)+\left(6{C}_{2}*4{C}_{2}\right)+\left(6{C}_{3}*4{C}_{1}\right)+6{C}_{2}$ = 209.

A) 215 | B) 315 |

C) 415 | D) 115 |

Explanation:

Hence, the given problem can be considered as selecting pairs of lines from the given 2 sets of parallel lines.

Therefore, the total number of parallelograms formed =

^{7}C

_{2}x

^{6}C

_{2}= 315

A) 340 | B) 370 |

C) 320 | D) 330 |

Explanation:

Here the order of choosing the elements doesn’t matter and this is a problem in combinations.

We have to find the number of ways of choosing 4 elements of this set which has 11 elements.

This can be done in 11${C}_{4}$ ways = 330 ways

A) 180 | B) 220 |

C) 240 | D) 160 |

Explanation:

Number of ways of choosing 2 black pens from 5 black pens in $5{C}_{2}$ ways.

Number of ways of choosing 2 white pens from 3 white pens in $3{C}_{2}$ ways.

Number of ways of choosing 2 red pens from 4 red pens in $4{C}_{2}$ways.

By the Counting Principle, 2 black pens, 2 white pens, and 2 red pens can be chosen in 10 x 3 x 6 =180 ways.

A) 2400 | B) 6400 |

C) 3600 | D) 7200 |

Explanation:

Let the beds be numbered 1 to 7.

**Case 1** : Suppose Anju is allotted bed number 1.

Then, Parvin cannot be allotted bed number 2.

So Parvin can be allotted a bed in 5 ways.

After alloting a bed to Parvin, the remaining 5 students can be allotted beds in 5! ways.

So, in this case the beds can be allotted in 5´5!ways = 600 ways.

**Case 2** : Anju is allotted bed number 7.

Then, Parvin cannot be allotted bed number 6

As in Case 1, the beds can be allotted in 600 ways.

**Case 3** : Anju is allotted one of the beds numbered 2,3,4,5 or 6.

Parvin cannot be allotted the beds on the right hand side and left hand side of Anju’s bed. For example, if Anju is allotted bed number 2, beds numbered 1 or 3 cannot be allotted to Parvin.

Therefore, Parvin can be allotted a bed in 4 ways in all these cases.

After allotting a bed to Parvin, the other 5 can be allotted a bed in 5! ways.

Therefore, in each of these cases, the beds can be allotted in 4´ 5! = 480 ways.

The beds can be allotted in (2x 600 + 5 x 480)ways = (1200 + 2400)ways = 3600 ways

A) 2520 | B) 2880 |

C) 4320 | D) 5040 |

Explanation:

The number of ways of arranging n beads in a necklace is $\frac{(n-1)!}{2}=\frac{(8-1)!}{2}=\frac{7!}{2}$ = **2520**

(since n = 8)