Permutations and CombinationsClass 11 Mathematics NCERT Solutions

Given: Digits available are 1, 2, 3, 4, and 5. We need to form 3-digit numbers.

Solution: We need to fill three places: Hundred's, Ten's, and Unit's.

By the fundamental principle of counting, the total number of 3-digit numbers is: $5 \times 5 \times 5 = 125$

Final Answer: 125 three-digit numbers can be formed if repetition is allowed.

Solution: We need to fill three places without repetition.

By the fundamental principle of counting, the total number of 3-digit numbers is: $5 \times 4 \times 3 = 60$

Final Answer: 60 three-digit numbers can be formed if repetition is not allowed.

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.

By the fundamental principle of counting, the total number of 3-digit even numbers is: $6 \times 6 \times 3 = 108$

Final Answer: 108 three-digit even numbers can be formed.

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 problem.

We need to fill 4 places:

By the fundamental principle of counting, the total number of codes is: $10 \times 9 \times 8 \times 7 = 5040$

Alternatively, using the permutation formula ${}^n P_r = \frac{n!}{(n-r)!}$: Here, $n=10$ and $r=4$. ${}^{10}P_4 = 10 \times 9 \times 8 \times 7 = 5040$

Final Answer: 5040 four-letter codes can be formed.

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 _ _ _

We need to fill the remaining 3 places. The digits 6 and 7 have been used. So, we are left with $10 - 2 = 8$ digits: {0, 1, 2, 3, 4, 5, 8, 9}.

By the fundamental principle of counting, the number of ways to fill the remaining 3 places is: $8 \times 7 \times 6 = 336$

Final Answer: 336 such telephone numbers can be constructed.

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).

By the fundamental principle of counting, the total number of possible outcomes for 3 tosses is: $2 \times 2 \times 2 = 8$

The possible outcomes are: {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}.

Final Answer: There are 8 possible outcomes.

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.

By the fundamental principle of counting, the total number of different signals is: $5 \times 4 = 20$

Alternatively, this is a permutation of 5 flags taken 2 at a time: ${}^5 P_2 = \frac{5!}{(5-2)!} = \frac{5!}{3!} = 5 \times 4 = 20$

Final Answer: 20 different signals can be generated.

Solution: By definition, $n! = n \times (n-1) \times \dots \times 2 \times 1$. $8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$ $8! = 40320$

Final Answer: $8! = 40320$.

Solution: First, calculate each factorial: $4! = 4 \times 3 \times 2 \times 1 = 24$ $3! = 3 \times 2 \times 1 = 6$

Now, subtract: $4! - 3! = 24 - 6 = 18$

Final Answer: $4! - 3! = 18$.

To Check: If $3! + 4! = 7!$.

Solution: Calculate the Left Hand Side (LHS): $3! = 3 \times 2 \times 1 = 6$ $4! = 4 \times 3 \times 2 \times 1 = 24$ $LHS = 3! + 4! = 6 + 24 = 30$

Calculate the Right Hand Side (RHS): $7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040$

Comparing LHS and RHS: $30 \neq 5040$ Therefore, $LHS \neq RHS$.

Final Answer: No, $3! + 4! \neq 7!$.

To Compute: $\frac{8!}{6! \times 2!}$

Solution: We can expand 8! until we get 6! to simplify the calculation. $8! = 8 \times 7 \times 6!$ $2! = 2 \times 1 = 2$

Now, substitute these into the expression: $\frac{8!}{6! \times 2!} = \frac{8 \times 7 \times 6!}{6! \times 2}$ Cancel out $6!$ from the numerator and denominator: $= \frac{8 \times 7}{2}$ $= \frac{56}{2}$ $= 28$

Final Answer: $\frac{8!}{6! \times 2!} = 28$.

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 $8!$: $8! \left( \frac{1}{6!} + \frac{1}{7!} \right) = 8! \left( \frac{x}{8!} \right)$ $\frac{8!}{6!} + \frac{8!}{7!} = x$

Now, simplify the terms: $\frac{8 \times 7 \times 6!}{6!} + \frac{8 \times 7!}{7!} = x$ $(8 \times 7) + 8 = x$ $56 + 8 = x$ $x = 64$

Final Answer: $x = 64$.

Formula: The expression $\frac{n!}{(n-r)!}$ is the formula for permutations, denoted as ${}^n P_r$.

Solution: Substitute $n=6$ and $r=2$ into the expression: $\frac{6!}{(6-2)!} = \frac{6!}{4!}$ Expand 6! until 4!: $= \frac{6 \times 5 \times 4!}{4!}$ Cancel out 4!: $= 6 \times 5 = 30$

Final Answer: For $n=6, r=2$, the value is 30.

Solution: Substitute $n=9$ and $r=5$ into the expression: $\frac{9!}{(9-5)!} = \frac{9!}{4!}$ Expand 9! until 4!: $= \frac{9 \times 8 \times 7 \times 6 \times 5 \times 4!}{4!}$ Cancel out 4!: $= 9 \times 8 \times 7 \times 6 \times 5$ $= 72 \times 42 \times 5$ $= 3024 \times 5 = 15120$

Final Answer: For $n=9, r=5$, the value is 15120.

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 allowed. We can use the permutation formula ${}^n P_r$. Here, $n=9$ and $r=3$.

Number of ways = ${}^9 P_3 = \frac{9!}{(9-3)!} = \frac{9!}{6!}$ $= 9 \times 8 \times 7$ $= 504$

Alternatively, using the fundamental principle of counting:

Final Answer: 504 three-digit numbers can be formed.

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.

By the fundamental principle of counting, the total number of 4-digit numbers is: $9 \times 9 \times 8 \times 7 = 4536$

Final Answer: There are 4536 four-digit numbers with no digit repeated.

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, starting with the unit's place due to the constraint.

After filling the unit's place, we are left with $6 - 1 = 5$ digits.

By the fundamental principle of counting, the total number of 3-digit even numbers is: $5 \times 4 \times 3 = 60$

Final Answer: 60 three-digit even numbers can be made.

Given: Digits available are {1, 2, 3, 4, 5}. Total 5 digits. No digit is repeated.

Solution: We need to arrange 4 digits out of 5. This is a permutation. Number of ways = ${}^5 P_4 = \frac{5!}{(5-4)!} = \frac{5!}{1!} = 5!$ $= 5 \times 4 \times 3 \times 2 = 120$

Final Answer for Part 1: 120 four-digit numbers can be formed.

Solution: For a number to be even, its unit's digit must be even. The available even digits are {2, 4}.

After filling the unit's place, we are left with 4 digits. We need to fill the remaining 3 places from these 4 digits. Number of ways to fill the remaining 3 places = ${}^4 P_3 = \frac{4!}{(4-3)!} = 4!$ $= 4 \times 3 \times 2 = 24$

Total number of even numbers = (Ways to fill remaining 3 places) $\times$ (Ways to fill unit's place) $= 24 \times 2 = 48$

Final Answer for Part 2: 48 of these numbers will be even.