ID: 160 Type: Mcq Source: NCERT

Question 1: The number of permutations of n different objects taken r at a time, where repetition is allowed, is

• A. \(n !\)
• B. \(r !\)
• C. \(n^r\)
• D. \(r^n\)

View Solution

✅ Manual Solution:

\(n^r\)

ID: 161 Type: Mcq

Question 2: The number of permutations of n objects, where p objects are of the same kind and rest are all different

• A. \( \frac{p!}{n!} \)
• B. \(n!\)
• C. \(p!\)
• D. \( \frac{n!}{p!} \)

View Solution View AI Solution

✅ Manual Solution:

\( \frac{n!}{p!} \)

🤖 AI Generated Solution:

The problem asks for the number of permutations of n objects where p objects are identical and the remaining n-p objects are distinct. **Step 1:** Consider the total number of permutations of n distinct objects. This is given by n!. **Step 2:** If p objects were distinct, they would contribute p! to the total permutations. However, since these p objects are identical, any arrangement among them does not lead to a new permutation. Therefore, we must divide the total permutations by the number of ways to arrange these p identical objects, which is p!. **Step 3:** The formula for permutations with repetitions is given by: $$ \text{Number of permutations} = \frac{n!}{n_1! n_2! \cdots n_k!} $$ where n is the total number of objects, and $n_1, n_2, \ldots, n_k$ are the counts of identical objects of each kind. **Step 4:** In this case, we have n objects in total. One group has p identical objects ($n_1 = p$). The remaining $n-p$ objects are all different, meaning each of them appears only once. So, we have $n-p$ groups of 1 object each. The counts for these are $1, 1, \ldots, 1$ (n-p times). **Step 5:** Applying the formula, the number of permutations is: $$ \frac{n!}{p! \times 1! \times 1! \times \cdots \times 1!} = \frac{n!}{p! \times 1} = \frac{n!}{p!} $$ The final answer is $\boxed{\frac{n!}{p!}}$.

ID: 146 Type: Very short (VSA) Source: NCERT

Question 3: How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that repetition of the digits is not allowed?

View Solution View AI Solution

✅ Manual Solution:

5*4*3=60

🤖 AI Generated Solution:

Here's how to solve this problem: **Understanding the Problem** We need to find the number of 3-digit numbers that can be formed using the digits 1, 2, 3, 4, and 5, without repeating any digit. This is a permutation problem because the order of the digits matters. **Step-by-step Solution** 1. **First digit:** We have 5 choices for the first digit (1, 2, 3, 4, or 5). 2. **Second digit:** Since repetition is not allowed, we have only 4 choices left for the second digit. 3. **Third digit:** Similarly, we have 3 choices remaining for the third digit. 4. **Total number of 3-digit numbers:** To find the total number of 3-digit numbers, we multiply the number of choices for each position. Number of 3-digit numbers = $5 \times 4 \times 3$ **Calculation** $5 \times 4 \times 3 = 60$ **Answer** There are 60 three-digit numbers that can be formed from the digits 1, 2, 3, 4, and 5 assuming that repetition of the digits is not allowed.

ID: 190 Type: Very short (VSA) Year: NCERT

Question 4: In how many ways can the letters of the word PERMUTATIONS be arranged if the vowels are all together

View Solution

✅ Manual Solution:

First note the letters in PERMUTATIONS: Total letters = 12 Vowels = E, U, A, I, O → 5 vowels (all distinct) Consonants = P, R, M, T, T, N, S → 7 consonants, with T repeated twice ✅ Step 1: Treat all vowels as one block Group the 5 vowels into a single unit: (E U A I O) Now instead of 12 separate letters, we have: 1 vowel block + 7 consonants = 8 total items But among the consonants, T repeats twice. Number of ways to arrange these 8 items: 8 ! 2 ! 2! 8! ​ (Since the two T's are identical.) ✅ Step 2: Arrange the 5 vowels within the block All 5 vowels are distinct, so they can be arranged among themselves in: 5 ! 5! ✅ Final Answer 8 ! 2 ! × 5 ! 2! 8! ​ ×5! ​ Now compute: 8 ! = 40320 8!=40320 2 ! = 2 2!=2 → 8 ! / 2 ! = 40320 / 2 = 20160 8!/2!=40320/2=20160 5 ! = 120 5!=120 So total arrangements: 20160 × 120 = 2419200 20160×120=2419200 ✅ Final Answer: 2 ,  ⁣ 419 ,  ⁣ 200 2,419,200 ​

ID: 145 Type: Short Source: NCERT

Question 5: How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that repetition of the digits is allowed?

ID: 147 Type: Short Source: NCERT

Question 6: Find the number of 4 letter words, with or without meaning, which can be formed out of the letters of the word ROSE, where the repetition of the letters is not allowed.

ID: 148 Type: Short Source: NCERT

Question 7: Given 4 flags of different colours, how many different signals can be generated, if a signal requires the use of 2 flags one below the other?

ID: 149 Type: Short Source: NCERT

Question 8: How many 2 digit even numbers can be formed from the digits 1, 2, 3, 4, 5 if the digits can be repeated?

ID: 150 Type: Short Source: NCERT

Question 9: Find the number of different signals that can be generated by arranging at least 2 flags in order (one below the other) on a vertical staff, if five different flags are available.

View AI Solution

🤖 AI Generated Solution:

**Correct Option if MCQ:** B **Reasoning:** * Number of signals with exactly 2 flags: $P(5, 2) = 5 \times 4 = 20$ * Number of signals with exactly 3 flags: $P(5, 3) = 5 \times 4 \times 3 = 60$ * Total signals: $20 + 60 + 120 + 120 = 320$

ID: 151 Type: Short Source: NCERT

Question 10: How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated?

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🤖 AI Generated Solution:

**Correct Option if MCQ:** Not applicable **Reasoning:** * Hundreds digit: 6 choices (1-6) * Tens digit: 6 choices (1-6) * Units digit: 3 choices (2, 4, 6) * Total numbers: $6 \times 6 \times 3 = 108$

ID: 152 Type: Short Source: NCERT

Question 11: How many 4-letter code can be formed using the first 10 letters of the English alphabet, if no letter can be repeated?

View AI Solution

🤖 AI Generated Solution:

**Correct Option if MCQ:** Not applicable **Reasoning:** * Permutations of 10 items taken 4 at a time. * $P(10, 4) = 10 \times 9 \times 8 \times 7$ * $10 \times 9 \times 8 \times 7 = 5040$

ID: 153 Type: Short Source: NCERT

Question 12: How many 5-digit telephone numbers can be constructed using the digits 0 to 9 if each number starts with 67 and no digit appears more than once?

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🤖 AI Generated Solution:

**Correct Option if MCQ:** Not applicable **Reasoning:** * The first two digits are fixed as 6 and 7. * Remaining 3 digits can be chosen from the remaining 8 digits (0-5, 8, 9) without repetition. * Number of permutations = $P(8, 3) = 8 \times 7 \times 6 = 336$.

ID: 154 Type: Short Source: NCERT

Question 13: A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there?

ID: 155 Type: Short Source: NCERT

Question 14: Evaluate the expression: \( \frac{n!}{r!(n-r)!} \) when \(n = 5\) and \(r = 2\).

ID: 156 Type: Short Source: NCERT

Question 15: If \(\frac{1}{8!} + \frac{1}{9!} = \frac{x}{10!}\), find x.

ID: 157 Type: Short Source: NCERT

Question 16: If \(\frac{1}{6!} + \frac{1}{6!} = \frac{x}{8!}\), find x.

ID: 158 Type: Short Source: NCERT

Question 17: Evaluate the expression: \( \frac{n!}{r!(n-r)!} \) when \(n = 6\) and \(r = 2\).

ID: 159 Type: Short Source: NCERT

Question 18: Evaluate the expression: \( \frac{n!}{r!(n-r)!} \) when \(n = 9\) and \(r = 5\).

ID: 162 Type: Short Source: NCERT

Question 19: Find the number of permutations of the letters of the word ALLAHABAD

View Solution

✅ Manual Solution:

7560

ID: 163 Type: Short Source: NCERT

Question 20: How many 4-digit numbers can be formed by using the digits 1 to 9 if Repetition of digits is not allowed?

View Solution

✅ Manual Solution:

\({}^9P_4 = \frac{9!}{(9-4)!} = \frac{9!}{5!} = 9 \times 8 \times 7 \times 6 = 3024.\)

ID: 165 Type: Short Source: NCERT

Question 21: Find the value of $n$ such that \({}^nP_5 = 42 \, {}^nP_3, \; n > 4\)

View Solution

✅ Manual Solution:

10

ID: 170 Type: Short Source: NCERT

Question 22: In how many ways can 4 red, 3 yellow and 2 green discs be arranged in a row if the discs of the same colour are indistinguishable ?

View Solution

✅ Manual Solution:

The number of arrangements \( \frac{9!}{4! 3! 2!}=1260\)

ID: 171 Type: Short Year: NCERT

Question 23: Find the number of arrangements of the letters of the word INDEPENDENCE.

View Solution

✅ Manual Solution:

The number of arrangements \( \frac{12!}{3! 4! 2!}=1663200\)

ID: 172 Type: Short Year: NCERT

Question 24: Find the number of arrangements of the letters of the word INDEPENDENCE.

View Solution

✅ Manual Solution:

The number of arrangements \( \frac{12!}{3! 4! 2!}=1663200\)

ID: 173 Type: Short Year: NCERT

Question 25: Find the number of arrangements of the letters of the word INDEPENDENCE so that the words start with P

View Solution

✅ Manual Solution:

The number of arrangements \( \frac{11!}{3! 4! 2!}=138600\)

ID: 178 Type: Short Year: NCERT

Question 26: How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated?

View Solution

✅ Manual Solution:

\(^9P_3=\)504

ID: 179 Type: Short Year: NCERT

Question 27: How many 4-digit numbers are there with no digit repeated?

View Solution

✅ Manual Solution:

4536

ID: 180 Type: Short Year: NCERT

Question 28: How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7, if no digit is repeated?

View Solution

✅ Manual Solution:

120

ID: 181 Type: Short Year: NCERT

Question 29: Find the number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, 5 if no digit is repeated. How many of these will be even?

View Solution

✅ Manual Solution:

48

ID: 182 Type: Short Year: NCERT

Question 30: From a committee of 8 persons, in how many ways can we choose a chairman and a vice chairman assuming one person can not hold more than one position?

View Solution

✅ Manual Solution:

56

ID: 183 Type: Short Year: NCERT

Question 31: How many words, with or without meaning, can be formed using all the letters of the word EQUATION, using each letter exactly once?

View Solution

✅ Manual Solution:

40320

ID: 184 Type: Short Year: NCERT

Question 32: How many words, with or without meaning can be made from the letters of the word MONDAY, assuming that no letter is repeated, if 4 letters are used at a time,

View Solution

✅ Manual Solution:

360

ID: 185 Type: Short Year: NCERT

Question 33: How many words, with or without meaning can be made from the letters of the word MONDAY, assuming that no letter is repeated, if all letters are used at a time

View Solution

✅ Manual Solution:

720

ID: 186 Type: Short Year: NCERT

Question 34: How many words, with or without meaning can be made from the letters of the word MONDAY, assuming that no letter is repeated, if all letters are used but first letter is a vowel?

View Solution

✅ Manual Solution:

240

ID: 187 Type: Short Year: NCERT

Question 35: In how many of the distinct permutations of the letters in MISSISSIPPI do the four I?s not come together?

View Solution

✅ Manual Solution:

33810

ID: 188 Type: Short Year: NCERT

Question 36: In how many ways can the letters of the word PERMUTATIONS be arranged if the words start with P and end with S

View Solution

✅ Manual Solution:

814400

ID: 189 Type: Short Year: NCERT

Question 37: In how many ways can the letters of the word PERMUTATIONS be arranged if the there are always 4 letters between P and S?

View Solution

✅ Manual Solution:

Let’s first think of possible placements: If P is in position 1, S will be in 6th place. If P is in 2, S will be in 7th, and so on. So positions for P–S with 4 letters between them can occur 7 ways (from 1–6, 2–7, …, 7–12). But since S can come before P as well, total ways to place the pair: 7×2=14 possible position pairs Now, fixing P and S, we have 10 letters remaining (E, R, M, U, T, A, T, I, O, N), which can be arranged: \(\dfrac{10!}{2!}=1,814,400\) Thus, total arrangements: 14×1,814,400=25,401,600

ID: 164 Type: Long Source: NCERT

Question 38: How many numbers lying between 100 and 1000 can be formed with the digits 0, 1, 2, 3, 4, 5, if the repetition of the digits is not allowed?

View Solution

✅ Manual Solution:

The required number\(= {}^{6}P_{3} - {}^{5}P_{2} = \frac{6!}{3!} - \frac{5!}{3!} = 4 \times 5 \times 6 - 4 \times 5 = 100\)

ID: 166 Type: Long Source: NCERT

Question 39: Find the value of \(n\) such that \(\dfrac{{}^nP_4}{{}^{n-1}P_4} = \dfrac{5}{3}, \; n > 4\)

View Solution

✅ Manual Solution:

10

ID: 167 Type: Long Source: NCERT

Question 40: Find \(r\), if \({}^5P_r = 6 \, {}^5P_{r-1}\).

View Solution

✅ Manual Solution:

8,3

ID: 168 Type: Long Source: NCERT

Question 41: Find the number of different 8-letter arrangements that can be made from the letters of the word DAUGHTER so that all vowels occur together

View Solution

✅ Manual Solution:

6 ! × 3 !

ID: 169 Type: Long Source: NCERT

Question 42: Find the number of different 8-letter arrangements that can be made from the letters of the word DAUGHTER so that all vowels do not occur together.

View Solution

✅ Manual Solution:

8 ! –( 6 ! × 3 !) =36000

ID: 174 Type: Long Year: NCERT

Question 43: Find the number of arrangements of the letters of the word INDEPENDENCE so that all the vowels always occur together

View Solution

✅ Manual Solution:

The number of arrangements \( \frac{8!}{3! 2!} ×\frac{5!}{ 4!}=16800\)

ID: 175 Type: Long Year: NCERT

Question 44: Find the number of arrangements of the letters of the word INDEPENDENCE so that the vowels never occur together.

View Solution

✅ Manual Solution:

The required number of arrangements = the total number of arrangements (without any restriction) – the number of arrangements where all the vowels occur together =\( \frac{12!}{3! 4! 2!}-\frac{8!5!}{3! 2! 4!}== 1663200 – 16800 = 1646400\)

ID: 176 Type: Long Year: NCERT

Question 45: Find the number of arrangements of the letters of the word INDEPENDENCE so that the vowels never occur together.

View Solution

✅ Manual Solution:

The required number of arrangements = the total number of arrangements (without any restriction) – the number of arrangements where all the vowels occur together =\( \frac{12!}{3! 4! 2!}-\frac{8!5!}{3! 2! 4!}== 1663200 – 16800 = 1646400\)

ID: 177 Type: Long Year: NCERT

Question 46: Find the number of arrangements of the letters of the word INDEPENDENCE so that the vowels never occur together.

View Solution

✅ Manual Solution:

The required number of arrangements = the total number of arrangements (without any restriction) – the number of arrangements where all the vowels occur together =\( \frac{12!}{3! 4! 2!}-\frac{8!5!}{3! 2! 4!}== 1663200 – 16800 = 1646400\)