CBSE Class 11 Mathematics Permutation And Combination Worksheet Set A - Free Printable

Problem: Counting Principle, Permutations, and Combinations Worksheet

#### 1. If a girl has 5 skirts, 8 shirts, and 6 pairs of shoes, how many outfits can she wear?

Solution:
To determine the total number of outfits, we use the counting principle, which states that if there are \( m \) ways to do one thing and \( n \) ways to do another, then there are \( m \times n \) ways to do both.

- Number of skirts: \( 5 \)
- Number of shirts: \( 8 \)
- Number of pairs of shoes: \( 6 \)

The total number of outfits is:
\[
5 \times 8 \times 6 = 240
\]

Answer: \( \boxed{240} \)

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#### 2. Three people run for class president, 4 for vice-president, and 2 for secretary. How many ways can the 3 officers be selected?

Solution:
This is a permutation problem because the order in which the officers are selected matters (president, vice-president, and secretary are distinct positions).

- Number of choices for president: \( 3 \)
- Number of choices for vice-president: \( 4 \)
- Number of choices for secretary: \( 2 \)

Using the counting principle:
\[
3 \times 4 \times 2 = 24
\]

Answer: \( \boxed{24} \)

---

#### 3. For a school lunch, you can get a hamburger, hot dog, or chicken sandwich. To drink, you can get a soda, juice, or milkshake. List all the different lunches consisting of 1 sandwich and 1 drink, and give the total number of possibilities as well.

Solution:
We need to list all possible combinations of sandwiches and drinks and then count them.

- Sandwich options: hamburger (H), hot dog (D), chicken (C) — 3 choices
- Drink options: soda (S), juice (J), milkshake (M) — 3 choices

The possible lunches are:
1. Hamburger with soda (H, S)
2. Hamburger with juice (H, J)
3. Hamburger with milkshake (H, M)
4. Hot dog with soda (D, S)
5. Hot dog with juice (D, J)
6. Hot dog with milkshake (D, M)
7. Chicken with soda (C, S)
8. Chicken with juice (C, J)
9. Chicken with milkshake (C, M)

There are \( 3 \) sandwiches and \( 3 \) drinks, so the total number of lunches is:
\[
3 \times 3 = 9
\]

Answer: \( \boxed{9} \)

---

#### 4. Sally has 7 candles, each a different color. How many ways can she arrange from left to right the candles in a candelabra that holds three candles?

Solution:
This is a permutation problem because the order in which the candles are arranged matters.

- Total candles: \( 7 \)
- Candles to be arranged: \( 3 \)

The number of ways to arrange 3 candles out of 7 is given by the permutation formula:
\[
P(n, r) = \frac{n!}{(n-r)!}
\]
where \( n = 7 \) and \( r = 3 \).

\[
P(7, 3) = \frac{7!}{(7-3)!} = \frac{7!}{4!} = \frac{7 \times 6 \times 5 \times 4!}{4!} = 7 \times 6 \times 5 = 210
\]

Answer: \( \boxed{210} \)

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#### 5. Eight toppings for pizza are available. In how many ways can Jim choose 3 of the toppings?

Solution:
This is a combination problem because the order in which the toppings are chosen does not matter.

- Total toppings: \( 8 \)
- Toppings to be chosen: \( 3 \)

The number of ways to choose 3 toppings out of 8 is given by the combination formula:
\[
C(n, r) = \binom{n}{r} = \frac{n!}{r!(n-r)!}
\]
where \( n = 8 \) and \( r = 3 \).

\[
C(8, 3) = \binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8!}{3! \cdot 5!} = \frac{8 \times 7 \times 6 \times 5!}{3! \cdot 5!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = \frac{336}{6} = 56
\]

Answer: \( \boxed{56} \)

---

#### 6. How many different ways can a committee of 3 students be selected from a class of 25?

Solution:
This is a combination problem because the order in which the students are selected does not matter.

- Total students: \( 25 \)
- Students to be selected: \( 3 \)

The number of ways to choose 3 students out of 25 is given by the combination formula:
\[
C(n, r) = \binom{n}{r} = \frac{n!}{r!(n-r)!}
\]
where \( n = 25 \) and \( r = 3 \).

\[
C(25, 3) = \binom{25}{3} = \frac{25!}{3!(25-3)!} = \frac{25!}{3! \cdot 22!} = \frac{25 \times 24 \times 23 \times 22!}{3! \cdot 22!} = \frac{25 \times 24 \times 23}{3 \times 2 \times 1} = \frac{13800}{6} = 2300
\]

Answer: \( \boxed{2300} \)

---

#### 7. A car dealer sells 4 makes of cars in 5 colors with either standard or automatic transmissions. How many variations does the dealership offer?

Solution:
We use the counting principle to determine the total number of variations.

- Number of makes: \( 4 \)
- Number of colors: \( 5 \)
- Number of transmission types: \( 2 \)

The total number of variations is:
\[
4 \times 5 \times 2 = 40
\]

Answer: \( \boxed{40} \)

---

#### 8. How many outcomes are possible when you roll 3 dice?

Solution:
Each die has 6 faces, and the outcome of each die is independent of the others. Using the counting principle:

- Number of outcomes for one die: \( 6 \)
- Number of dice: \( 3 \)

The total number of outcomes is:
\[
6 \times 6 \times 6 = 6^3 = 216
\]

Answer: \( \boxed{216} \)

---

#### 9. A briefcase lock has 3 rotating cylinders, each containing 10 digits. How many numerical codes are possible?

Solution:
Each cylinder has 10 digits (0 through 9), and the digits on each cylinder are independent of the others. Using the counting principle:

- Number of digits per cylinder: \( 10 \)
- Number of cylinders: \( 3 \)

The total number of numerical codes is:
\[
10 \times 10 \times 10 = 10^3 = 1000
\]

Answer: \( \boxed{1000} \)

---

#### 10. A golf club manufacturer makes irons with 7 different shaft lengths, 3 different grips, and 2 different club head materials. How many different combinations are offered?

Solution:
We use the counting principle to determine the total number of combinations.

- Number of shaft lengths: \( 7 \)
- Number of grips: \( 3 \)
- Number of club head materials: \( 2 \)

The total number of combinations is:
\[
7 \times 3 \times 2 = 42
\]

Answer: \( \boxed{42} \)

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#### 11. How many 7-digit telephone numbers can be formed if the first digit cannot be 0 or 1?

Solution:
For a 7-digit telephone number:
- The first digit cannot be \( 0 \) or \( 1 \), so it has \( 8 \) possible values (\( 2, 3, 4, 5, 6, 7, 8, 9 \)).
- Each of the remaining 6 digits can be any digit from \( 0 \) to \( 9 \), so each has \( 10 \) possible values.

Using the counting principle:
\[
8 \times 10 \times 10 \times 10 \times 10 \times 10 \times 10 = 8 \times 10^6 = 8,000,000
\]

Answer: \( \boxed{8000000} \)

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Final Answers:

1. \( \boxed{240} \)
2. \( \boxed{24} \)
3. \( \boxed{9} \)
4. \( \boxed{210} \)
5. \( \boxed{56} \)
6. \( \boxed{2300} \)
7. \( \boxed{40} \)
8. \( \boxed{216} \)
9. \( \boxed{1000} \)
10. \( \boxed{42} \)
11. \( \boxed{8000000} \)