# 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) 7600 | B) 7200 |

C) 6400 | D) 3600 |

Explanation:

From 5 consonants, 3 consonants can be selected in $5{C}_{3}$ ways.

From 4 vowels, 2 vowels can be selected in $4{C}_{2}$ ways.

Now with every selection, number of ways of arranging 5 letters is $5{P}_{5}$ways.

Total number of words = $5{C}_{3}*4{C}_{2}*5{P}_{5}$

= 10x 6 x 5 x 4 x 3 x 2 x 1= 7200

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

A) 18! | B) 18! x 19! |

C) 18!(6 x 24) | D) 18! x 30 |

Explanation:

Here clock-wise and anti-clockwise arrangements are same.

Hence total number of circular–permutations: $\frac{18{P}_{12}}{2*12}$ = $\frac{18!}{6*24}$

A) 36 | B) 25 |

C) 24 | D) 72 |

Explanation:

The toys are different; The boxes are identical

If none of the boxes is to remain empty, then we can pack the toys in one of the following ways

a. 2, 2, 1

b. 3, 1, 1

**Case a.** Number of ways of achieving the first option 2 - 2 - 1

Two toys out of the 5 can be selected in $5{C}_{2}$ ways. Another 2 out of the remaining 3 can be selected in $3{C}_{2}$ ways and the last toy can be selected in $1{C}_{1}$ way.

However, as the boxes are identical, the two different ways of selecting which box holds the first two toys and which one holds the second set of two toys will look the same. Hence, we need to divide the result by 2

Therefore, total number of ways of achieving the 2 - 2 - 1 option is ways $5{C}_{2}*3{C}_{2}$= 15 ways

**Case b**. Number of ways of achieving the second option 3 - 1 - 1

Three toys out of the 5 can be selected in $5{C}_{3}$ ways. As the boxes are identical, the remaining two toys can go into the two identical looking boxes in only one way.

Therefore, total number of ways of getting the 3 - 1 - 1 option is $5{C}_{3}$ = 10 = 10 ways.

Total ways in which the 5 toys can be packed in 3 identical boxes

= number of ways of achieving Case a + number of ways of achieving Case b= 15 + 10 = 25 ways.

A) 18000 | B) 17280 |

C) 17829 | D) 18270 |

Explanation:

Let 4 girls be one unit and now there are 6 units in all.

They can be arranged in 6! ways.

In each of these arrangements 4 girls can be arranged in 4! ways.

Total number of arrangements in which girls are always together = 6! x 4!= 720 x 24 = 17280

A) 1170 | B) 1200 |

C) 720 | D) 360 |

Explanation:

Possibilities Bowlers Batsmen Number of ways

6 9

1 4 7 $(6{C}_{4}*9{C}_{7})$

2 5 6 $\left(6{C}_{5}*9{C}_{6}\right)$

3 6 5 $\left(6{C}_{6}*9{C}_{5}\right)$

$\left(6{C}_{4}*9{C}_{7}\right)$ = 15 x 36 = 540

$\left(6{C}_{5}*9{C}_{6}\right)$ = 6 x 84 = 504

$\left(6{C}_{6}*9{C}_{5}\right)$= 1 x 126 = 126

Total = 1170

A) 569 | B) 729 |

C) 625 | D) 769 |

Explanation:

Choose 5 starters from a team of 12 players. Order is not important.

$12{C}_{5}$= 729

A) 126 | B) 120 |

C) 146 | D) 156 |

Explanation:

There are 8 students and the maximum capacity of the cars together is 9.

We may divide the 8 students as follows

Case I: 5 students in the first car and 3 in the second

Case II: 4 students in the first car and 4 in the second

Hence, in Case I: 8 students are divided into groups of 5 and 3 in8C3 ways.

Similarly, in Case II: 8 students are divided into two groups of 4 and 4 in 8C4ways.

Therefore, the total number of ways in which 8 students can travel is:

$8{C}_{3}+8{C}_{4}$=56 + 70= 126