Question 1

Q1: How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that (i) repetition of the digits is allowed? (ii) repetition of the...

Accepted Answer

Given: Digits available are 1, 2, 3, 4, and 5. We need to form 3-digit numbers. (i) Repetition of the digits is allowed Solution: We need to fill three places: Hundred's, Ten's, and Unit's. - The hundred's place can be filled in 5 ways (using any of the 5 digits). - Since repetition is allowed, the...

Question 2

Q2: How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated?

Accepted Answer

Given: Digits available are 1, 2, 3, 4, 5, 6. We need to form 3-digit even numbers, and repetition is allowed. To Find: The number of possible 3-digit even numbers. Solution: For a number to be even, its unit's digit must be an even number. - The available even digits are 2, 4, and 6. So, the unit's...

Question 3

Q3: How many 4-letter code can be formed using the first 10 letters of the English alphabet, if no letter can be repeated?

Accepted Answer

Given: The first 10 letters of the English alphabet. We need to form a 4-letter code, and no letter can be repeated. To Find: The number of possible 4-letter codes. Solution: This is a problem of arranging 4 letters out of 10, where order matters and repetition is not allowed. This is a permutation...

Question 4

Q4: 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?

Accepted Answer

Given: Digits available are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. We need to form a 5-digit telephone number that starts with 67, and no digit is repeated. To Find: The number of possible telephone numbers. Solution: The 5-digit number has the form: 6 7 _ _ _ - The first digit is fixed as 6. - The second di...

Question 5

Q5: A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there?

Accepted Answer

Given: A coin is tossed 3 times. To Find: The total number of possible outcomes. Solution: When a coin is tossed once, there are 2 possible outcomes: Head (H) or Tail (T). - For the first toss, there are 2 possible outcomes. - For the second toss, there are 2 possible outcomes. - For the third toss,...

Question 6

Q6: Given 5 flags of different colours, how many different signals can be generated if each signal requires the use of 2 flags, one below the other?

Accepted Answer

Given: 5 flags of different colours. A signal requires 2 flags, one below the other. To Find: The number of different signals that can be generated. Solution: A signal consists of an upper flag and a lower flag. The order matters. - The position for the upper flag can be filled in 5 ways (using any...

Question 7

Q1: Evaluate (i) 8! (ii) 4! - 3!

Accepted Answer

(i) Evaluate 8! Solution: By definition, $n! = n 	imes (n-1) 	imes \dots 	imes 2 	imes 1$. $$8! = 8 	imes 7 	imes 6 	imes 5 	imes 4 	imes 3 	imes 2 	imes 1$$ $$8! = 40320$$ Final Answer: $8! = 40320$. (ii) Evaluate 4! - 3! Solution: First, calculate each factorial: $$4! = 4 	imes 3 	ime...

Question 8

Q2: Is 3! + 4! = 7! ?

Accepted Answer

To Check: If $3! + 4! = 7!$. Solution: Calculate the Left Hand Side (LHS): $$3! = 3 	imes 2 	imes 1 = 6$$ $$4! = 4 	imes 3 	imes 2 	imes 1 = 24$$ $$LHS = 3! + 4! = 6 + 24 = 30$$ Calculate the Right Hand Side (RHS): $$7! = 7 	imes 6 	imes 5 	imes 4 	imes 3 	imes 2 	imes 1 = 5040$$ Comparin...

Question 9

Q3: Compute $\frac{8!}{6! 	imes 2!}$

Accepted Answer

To Compute: $\frac{8!}{6! 	imes 2!}$ Solution: We can expand 8! until we get 6! to simplify the calculation. $$8! = 8 	imes 7 	imes 6!$$ $$2! = 2 	imes 1 = 2$$ Now, substitute these into the expression: $$\frac{8!}{6! 	imes 2!} = \frac{8 	imes 7 	imes 6!}{6! 	imes 2}$$ Cancel out $6!$ from t...

Question 10

Q4: If $\frac{1}{6!} + \frac{1}{7!} = \frac{x}{8!}$, find $x$

Accepted Answer

Given: The equation $\frac{1}{6!} + \frac{1}{7!} = \frac{x}{8!}$ To Find: The value of $x$. Solution: To solve for $x$, we can make the denominators common or multiply the entire equation by the largest factorial, which is $8!$. Multiplying both sides by $8!$: $$8! \left( \frac{1}{6!} + \frac{1}{7!}...

Question 11

Q5: Evaluate $\frac{n!}{(n-r)!}$, when (i) n = 6, r = 2 (ii) n = 9, r = 5.

Accepted Answer

Formula: The expression $\frac{n!}{(n-r)!}$ is the formula for permutations, denoted as ${}^n P_r$. (i) n = 6, r = 2 Solution: Substitute $n=6$ and $r=2$ into the expression: $$\frac{6!}{(6-2)!} = \frac{6!}{4!}$$ Expand 6! until 4!: $$= \frac{6 	imes 5 	imes 4!}{4!}$$ Cancel out 4!: $$= 6 	imes 5...

Question 12

Q1: How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated?

Accepted Answer

Given: Digits available are {1, 2, 3, 4, 5, 6, 7, 8, 9}. Total 9 digits. We need to form 3-digit numbers with no repetition. To Find: The total number of such 3-digit numbers. Solution: This is a problem of arranging 3 digits out of 9 available digits, where order matters and repetition is not allow...

Question 13

Q2: How many 4-digit numbers are there with no digit repeated?

Accepted Answer

Given: Digits available are {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. Total 10 digits. We need to form 4-digit numbers with no repetition. To Find: The total number of such 4-digit numbers. Solution: We need to fill four places: Thousands, Hundreds, Tens, and Units. - The thousands place cannot be 0, so it ca...

Question 14

Q3: How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7, if no digit is repeated?

Accepted Answer

Given: Digits available are {1, 2, 3, 4, 6, 7}. Total 6 digits. We need to form 3-digit even numbers with no repetition. To Find: The total number of such numbers. Solution: For a number to be even, its unit's digit must be even. The available even digits are {2, 4, 6}. Let's fill the places, starti...

Question 15

Q4: 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?

Accepted Answer

Given: Digits available are {1, 2, 3, 4, 5}. Total 5 digits. No digit is repeated. Part 1: Total number of 4-digit numbers Solution: We need to arrange 4 digits out of 5. This is a permutation. Number of ways = ${}^5 P_4 = \frac{5!}{(5-4)!} = \frac{5!}{1!} = 5!$ $$= 5 	imes 4 	imes 3 	imes 2 = 12...

NCERT Solutions