# Permutation and Combination (PnC) – Important Concepts for Placement Aptitude

#### 1. Fundamental Rule

If one operation can be performed in ‘m’ ways and (when, it has been performed in any one of these ways), a second operation then can be performed in ‘n’ ways, the number of ways of performing the two operations will be m x n.

**Q: If there are three cities A, B and C such that there are 3 roads connecting A and B and 4 roads connecting B and C, then in how many ways one can travel from A to C?**

Sol: 3 x 4 = 12

#### 2. Permutations

- A permutation is an arrangement of objects in a specific order.
- The number of permutations of n things taking r at a time is denoted by ” \( ^{n}P_r \)” (read as “nPr”) and can be calculated as $$ ^{n}P_r = \frac{n!} {(n-r)!}$$ where n! denotes the factorial of n, which is the product of all positive integers less than or equal to n

**Q: In how many different ways can 5 friends sit in a row of 5 seats?**

Sol: 5!=5×4×3×2×1=120

**Q: A lock has 4 different dials, each labeled with digits from 0 to 9. How many different 4-digit lock codes can be formed if no digit can be repeated?**

Sol: P(10,4)=10×9×8×7=5040

**Q: You have 6 different books. In how many ways can you arrange 3 of them on a shelf?**

Sol: The arrangement matters, so it’s a permutation problem

The number of ways to arrange 3 books out of 6 is P(6,3)=6×5×4=120

**Q: There are 10 participants in a race. In how many different ways can gold, silver, and bronze medals be awarded?**

Sol: The order of the medals (gold, silver, bronze) matters, so it’s a permutation.

The number of ways to distribute 3 medals to 10 participants is: P(10,3)=10×9×8=720

**Q: A password requires exactly 5 characters from a set of 10 unique characters. How many different 5-character passwords can be formed if no character is repeated?**

Sol: The order of the characters matters, so it’s a permutation problem.

The number of ways to arrange 5 characters out of 10 is: P(10,5)=10×9×8×7×6=30,240

#### 3. Combination

- A combination is a selection of objects without regard to the order.
- The number of combinations of n things taking r at time is denoted by \( ^{n}C_r (read as nCr) and calculated as $$^{n}C_r = \frac{n!}{r! \cdot (n-r)!} $$ where n! denotes the factorial of n, which is the product of all positive integers less than or equal to n.

**Q: You have 8 types of fruit and want to select 3 different fruits to make a fruit salad. How many different ways can you choose the fruits?**

Sol: The order doesn’t matter here, so it’s a combination problem. The number of ways to choose 3 fruits out of 8 is: C(8,3)=56

**Q: A group consists of 7 boys and 5 girls. If you want to form a committee of 4 people, how many different ways can you choose the committee members?**

Sol: The number of ways to choose 4 people out of 12 (7 boys + 5 girls) is: C(12,4) = 495

**Q: There are 10 players in a basketball team. The coach wants to choose 5 players to start the game. How many different ways can the coach choose the starting 5 players?**

Sol: The order doesn’t matter, so it’s a combination problem. The number of ways to choose 5 players out of 10 is: C(10,5) = 252

**Q: A florist has 12 different types of flowers and wants to create a bouquet with 4 different types of flowers. How many different ways can the florist choose the flowers?**

Sol: The order doesn’t matter here, so it’s a combination problem. The number of ways to choose 4 flowers out of 12 is: C(12,4) = 495

**Q: Out of 6 available movies, you want to watch 2 over the weekend. In how many ways can you select the two movies?**

Sol: Since the order doesn’t matter (you’re just picking movies to watch), this is a combination. The number of ways to choose 2 movies out of 6 is: C(6,2)=15

#### 4. Special cases

**CASE 1 : Number of arrangements of n items of which p are of one type, q are of a second type and the rest are distinct –**

The number of ways in which n things may be arranged taking them all at a time, when p of the things are exactly like one kind, q of them exactly like another kind, r of them exactly like of a third kind, and the rest all distinct is $$ \frac{n!}{(p! \ q! \ r!)} $$

**Q: You have 7 letters: A, A, A, B, B, C, D. In how many distinct ways can you arrange these letters?**

Sol: We have 7 letters in total. There are 3 A’s (identical), 2 B’s (identical), and 1 C and 1 D (distinct)

Number of arrangements = $\frac{7!}{3! \times 2! \times 1! \times 1!}$ = 420

**CASE 2 : Number of arrangements of n distinct items where each item can be used any number of times (i.e., repetition allowed)**

- When selecting items with repetition allowed, each selection can be made independently and has the same number of choices, which is n.
- For r selections, each selection can be made in n ways, resulting in a total of \( n \times n \times n \) … (r times) ways, which simplifies to \( n^r \).
- Thus, the number of permutations of n things taken r at a time with repetition allowed is \( \longrightarrow n^r \).
- Note – When solving questions of such type it is better to grasp the reasoning behind the formula rather than memorizing it directly.

**Q: You want to create a 4-digit passcode using the numbers 1, 2, 3, 4, and 5. Each digit can be used more than once. How many different passcodes can you form?**

Sol: We have 5 distinct digits (1, 2, 3, 4, 5). The length of the passcode is 4 digits. Since repetition is allowed, each digit can appear multiple times.

The formula for the number of permutations of n distinct items taken r at a time with repetition allowed is: = $ n^r = 5^4$ = 625

#### 5. Circular Permutations

- In a straight line, arranging n items results in n! permutations, but in a circle, due to rotation, shifting each item by one place doesn’t change the arrangement.
- In a circular arrangement, after fixing one item, the remaining (n-1) items can be arranged in (n-1)! ways.
- Hence, the number of ways in which n distinct things can be arranged in a circular arrangement is $$(n-1)!$$
- The number of circular arrangements of n distinct items is –
- (n-1)! \( \longrightarrow \) if there is difference between clockwise and anticlockwise arrangements
- (n-1)!/2 \( \longrightarrow \) if there is NO DIFFERENCE between clockwise and anticlockwise arrangements

**Q: You have 5 distinct flowers, and you want to arrange them in a circular pattern for a decoration. How many different ways can you arrange these flowers if clockwise and anticlockwise arrangements are considered different?**

Sol: Since clockwise and anticlockwise arrangements are considered different, the formula for the number of arrangements is : (n−1)!=(5−1)!=4!=4×3×2×1=24

**Q: You have 6 friends, and you want to arrange them in a circular seating arrangement at a round table. There is no difference between clockwise and anticlockwise seating arrangements (i.e., the arrangement is the same if everyone shifts their position clockwise or anticlockwise). How many distinct seating arrangements can be made?**

Sol: Since there is no difference between clockwise and anticlockwise arrangements, we divide the number of circular arrangements by 2. The formula is = $\frac{(n-1)!}{2} = \frac{(6-1)!}{2} = \frac{5!}{2} = 60$

#### 6. Rank of a word

The rank of a word is its position when all possible words formed by arranging its letters once are sorted alphabetically.

**Q: What is the rank of the word “POINT” in dictionary among all the possible words that can be formed by rearranging the letters?**

Sol: Consider the word “POINT”. Arrange letters in alphabetical order: I, N, O, P, T. Count words preceding “POINT” alphabetically. Words beginning with I: 4! , Words beginning with N: 4! , Words beginning with O: 4! , Words beginning with PI: 3! , Words beginning with PN: 3! . then we arrive at the word ‘POINT’. \( \\ \) There are \( 3 \times 4!+2 \times 3!= 84 \ words\) that precede the word POINT i.e., POINT is the 85th word. \( \\ \) Hence rank of ‘POINT’ is 85.

#### 7. The number of diagonals in an n-sided regular polygon

- An n-sided regular polygon has n vertices. Joining any two vertices we get a line of the polygon which are \( ^{n}C_2 \) in number: Of these \( ^{n}C_2 \) lines, n of them are sides.
- Hence diagonals are $$^{n}C_{2}-n=\frac{n(n-3)}{2} $$

You can also refer following videos to grasp basics of percentage in more details

Refer Topic: Ratio and Proportion: https://www.learntheta.com/placement-aptitude-ratio-proportion/

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