GSEB Class 11 Maths Solutions Chapter 7 Permutations and Combinations Exercise 7.2

Question 1. Evaluate:
(i) \( 8! \)
(ii) \( 4! - 3! \)
Answer:
(i) To calculate \( 8! \), we multiply all positive integers from 8 down to 1:
\( 8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40320 \).
The factorial of 8 is 40320.
(ii) To evaluate \( 4! - 3! \), we first calculate each factorial separately:
\( 4! = 4 \times 3 \times 2 \times 1 = 24 \)
\( 3! = 3 \times 2 \times 1 = 6 \)
Now, we subtract the values:
\( 4! - 3! = 24 - 6 = 18 \).
In simple words: For part (i), you multiply all numbers from 8 down to 1 to find its factorial. For part (ii), calculate 4 factorial and 3 factorial separately, then subtract the second result from the first to get the final answer.

Exam Tip: Remember that \( n! \) means multiplying all whole numbers from \( n \) down to 1. Always evaluate each factorial completely before performing addition, subtraction, or other operations.

Question 2. Verify whether \( 3! + 4! = 7! \).
Answer:
First, we calculate the values of each factorial involved in the equation:
\( 3! = 3 \times 2 \times 1 = 6 \)
\( 4! = 4 \times 3 \times 2 \times 1 = 24 \)
\( 7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040 \)
Next, we evaluate the left-hand side (L.H.S.) of the equation:
\( \text{L.H.S.} = 3! + 4! = 6 + 24 = 30 \)
Now, we compare this value with the right-hand side (R.H.S.) of the equation:
\( \text{R.H.S.} = 7! = 5040 \)
Since \( 30 \ne 5040 \), we can confirm that \( 3! + 4! \ne 7! \).
In simple words: We calculated the factorials for 3, 4, and 7. Adding 3 factorial (6) and 4 factorial (24) gives 30. This is not the same as 7 factorial (5040). So, the statement is false.

Exam Tip: Factorials are not distributive over addition. Always calculate each factorial value first before performing any arithmetic operations like addition or subtraction.

Question 3. Compute \( \frac{8!}{6! \times 2!} \).
Answer:
To compute the expression \( \frac{8!}{6! \times 2!} \), we can expand the factorials and simplify.
We know that \( 8! = 8 \times 7 \times 6! \).
And \( 2! = 2 \times 1 = 2 \).
Substitute these values 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:
\( = \frac{8 \times 7}{2!} \)
\( = \frac{8 \times 7}{2 \times 1} \)
\( = \frac{56}{2} \)
\( = 28 \)
Therefore, the value of the expression is 28.
In simple words: To work this out, write 8 factorial as \( 8 \times 7 \times 6! \). Then you can cancel out \( 6! \) from the top and bottom. What's left is \( (8 \times 7) \) divided by \( 2! \) (which is 2). This gives \( 56 \) divided by \( 2 \), resulting in 28.

Exam Tip: When dealing with ratios of factorials, expand the larger factorial until it matches the smaller one in the denominator. This allows for quick cancellation and simplification.

Question 4. If \( \frac{1}{6!} + \frac{1}{7!} = \frac{x}{8!} \), find x?
Answer:
We need to find the value of \( x \) in the equation \( \frac{1}{6!} + \frac{1}{7!} = \frac{x}{8!} \).
First, let's combine the terms on the left-hand side. To do this, we express \( \frac{1}{6!} \) with a denominator of \( 7! \).
We know that \( 7! = 7 \times 6! \), so \( 6! = \frac{7!}{7} \).
\( \frac{1}{6!} = \frac{1 \times 7}{6! \times 7} = \frac{7}{7!} \)
Now, substitute this back into the equation:
\( \frac{7}{7!} + \frac{1}{7!} = \frac{x}{8!} \)
Add the fractions on the left side:
\( \frac{7+1}{7!} = \frac{x}{8!} \)
\( \frac{8}{7!} = \frac{x}{8!} \)
To solve for \( x \), multiply both sides by \( 8! \):
\( x = \frac{8}{7!} \times 8! \)
We know that \( 8! = 8 \times 7! \). Substitute this into the equation:
\( x = \frac{8}{7!} \times (8 \times 7!) \)
Cancel out \( 7! \) from the numerator and denominator:
\( x = 8 \times 8 \)
\( x = 64 \)
Thus, the value of \( x \) is 64.
In simple words: To find \( x \), first make the denominators the same on the left side by writing \( 1/6! \) as \( 7/7! \). Add the fractions to get \( 8/7! \). Now you have \( 8/7! = x/8! \). To get \( x \), multiply both sides by \( 8! \). Since \( 8! \) is \( 8 \times 7! \), you can cancel \( 7! \) and are left with \( 8 \times 8 \), which equals 64.

Exam Tip: When adding or subtracting fractions with factorials, convert all terms to have the largest factorial in the denominator to simplify calculations effectively.

Question 5. Evaluate \( \frac{n!}{(n-r)!} \), when:
(i) \( n = 6, r = 2 \)
(ii) \( n = 9, r = 5 \)
Answer:
We need to evaluate the expression \( \frac{n!}{(n-r)!} \) for the given values of \( n \) and \( r \).
(i) When \( n = 6, r = 2 \):
Substitute the values into the expression:
\( \frac{6!}{(6-2)!} = \frac{6!}{4!} \)
Expand \( 6! \) until \( 4! \):
\( = \frac{6 \times 5 \times 4!}{4!} \)
Cancel out \( 4! \) from the numerator and denominator:
\( = 6 \times 5 \)
\( = 30 \)
So, for \( n = 6, r = 2 \), the value is 30.
(ii) When \( n = 9, r = 5 \):
Substitute the values 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 \( 4! \) from the numerator and denominator:
\( = 9 \times 8 \times 7 \times 6 \times 5 \)
\( = 72 \times 42 \times 5 \)
\( = 72 \times 210 \)
\( = 15120 \)
So, for \( n = 9, r = 5 \), the value is 15120.
In simple words: For each part, put the given \( n \) and \( r \) numbers into the formula \( n! / (n-r)! \). Then, expand the factorial on top until you can cancel out the factorial on the bottom. Multiply the remaining numbers to get the answer.

Exam Tip: This specific formula, \( \frac{n!}{(n-r)!} \), represents the number of permutations of \( n \) distinct items taken \( r \) at a time. It's often denoted as \( P(n, r) \) or \( _nP_r \).