GSEB Class 11 Maths Solutions Chapter 7 ક્રમચય અને સંચય Exercise 7.2

Question 1. કિંમત શોધો : (1)8! (2) 4! – 3!
Answer:
(1) \( 8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40320 \)
(2) \( 4! - 3! = (4 \times 3 \times 2 \times 1) - (3 \times 2 \times 1) \)
\( = 24 - 6 \)
\( = 18 \)
In simple words: First, calculate the factorial for each number. Then, for the second part, subtract the result of 3! from 4! to get the final answer.

Exam Tip: Remember that \( n! \) represents the product of all positive integers up to \( n \). Be careful with the order of operations when both factorials and subtraction are involved.

Question 2. \( 3! + 4! = 7! \) થશે કે નહિ તે નક્કી કરો.
Answer: ના, કારણ કે
\( 3! + 4! = (3 \times 2 \times 1) + (4 \times 3 \times 2 \times 1) \)
\( = 6 + 24 \)
\( = 30 \)
\( 7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \)
\( = 5040 \)
\( \implies 3! + 4! \neq 7! \)
In simple words: First, work out the value of 3! plus 4!. Then, find the value of 7!. If the two values are not the same, then the statement is false.

Exam Tip: Factorials do not distribute over addition; that is, \( (a+b)! \neq a! + b! \). Always calculate each factorial separately before performing addition or subtraction.

Question 3. કિંમત શોધો : \( \frac{8 !}{6 ! \times 2 !} \)
Answer:
\( \frac{8 !}{6 ! \times 2 !} = \frac{8 \times 7 \times 6 !}{6 ! \times 2 \times 1} \)
\( = \frac{8 \times 7}{2 \times 1} \)
\( = \frac{56}{2} \)
\( = 28 \)
In simple words: To simplify this fraction, expand the top factorial (8!) until you reach 6!, then cancel out the common 6! terms from the top and bottom. Finish the calculation by dividing by 2!.

Exam Tip: When simplifying fractions with factorials, expand the larger factorial until it matches the smaller one in the denominator to allow for easy cancellation. This makes calculations simpler.

Question 4. જો \( \frac{1}{6!}+\frac{1}{7 !}=\frac{x}{8 !} \) હોય, તો \( x \) ની કિંમત શોધો.
Answer:
\( \frac{1}{6!}+\frac{1}{7 !}=\frac{x}{8 !} \)
\( \implies \frac{1}{6!}+\frac{1}{7 \times 6!}=\frac{x}{8 \times 7 \times 6!} \)
\( \implies \frac{1}{6!} \left(1+\frac{1}{7}\right)=\frac{x}{8 \times 7 \times 6!} \)
\( \implies \frac{1}{6!} \left(\frac{7+1}{7}\right)=\frac{x}{56 \times 6!} \)
\( \implies \frac{1}{6!} \times \frac{8}{7}=\frac{x}{56 \times 6!} \)
Multiply both sides by \( 56 \times 6! \):
\( \implies 56 \times 6! \times \frac{1}{6!} \times \frac{8}{7} = x \)
\( \implies 56 \times \frac{8}{7} = x \)
\( \implies 8 \times 8 = x \)
\( \implies x = 64 \)
In simple words: To find \( x \), first rewrite the fractions so they have a common 6! term. Then, combine the fractions on the left side and cancel out the 6! from both sides. Solve for \( x \) by multiplying and simplifying.

Exam Tip: When solving equations with factorials, it is often helpful to expand the larger factorials to reveal common smaller factorials. This allows for simplification and cancellation, making the equation easier to solve.

Question 5. જ્યારે (1)n = 6, r = 2 (2)n = 9, r = 5 હોય ત્યારે \( \frac{n !}{(n-r) !} \) ની કિંમત શોધો.
Answer:
(1) \( n = 6, r = 2 \)
\( \frac{n!}{(n-r)!} = \frac{6!}{(6-2)!} \)
\( = \frac{6!}{4!} \)
\( = \frac{6 \times 5 \times 4!}{4!} \)
\( = 6 \times 5 \)
\( = 30 \)
(2) \( n = 9, r = 5 \)
\( \frac{n!}{(n-r)!} = \frac{9!}{(9-5)!} \)
\( = \frac{9!}{4!} \)
\( = \frac{9 \times 8 \times 7 \times 6 \times 5 \times 4!}{4!} \)
\( = 9 \times 8 \times 7 \times 6 \times 5 \)
\( = 15120 \)
In simple words: To calculate this value, first subtract \( r \) from \( n \) inside the factorial. Then, expand the top factorial (\( n! \)) until you reach the factorial of \( (n-r)! \), which lets you cancel out the common terms and multiply the remaining numbers.

Exam Tip: This formula represents the number of permutations \( P(n,r) \). Always simplify by expanding the numerator's factorial down to the denominator's factorial to avoid calculating very large numbers unnecessarily.