Question 1, Exercise 6.1
Solutions of Question 1 of Exercise 6.1 of Unit 06: Permutation and Combination. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.
Question 1(i)
Evaluate $10!$.
Solution.
\begin{align*} 10! &= 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \\ &= 3628800 \end{align*}
Question 1(ii)
Evaluate $\dfrac{12!}{7! 3! 2!}$.
Solution.
\begin{align*} \dfrac{12!}{7! \, 3! \, 2!} &= \dfrac{12 \times 11 \times 10 \times 9 \times 8 \times 7!}{7! \times (3 \times 2 \times 1) \times (2 \times 1)} \\ &= \dfrac{12 \times 11 \times 10 \times 9 \times 8}{3 \times 2 \times 1 \times 2 \times 1} \\ &= \dfrac{12 \times 11 \times 10 \times 9 \times 8}{12} \\ &= 11 \times 10 \times 9 \times 8 \\ &= 7920 \end{align*}
Question 1(iii)
Evaluate $\dfrac{4!-2!}{3!+5!}$
Solution.
\begin{align*} \dfrac{4! - 2!}{3! + 5!} &= \dfrac{(4 \times 3 \times 2 \times 1) - (2 \times 1)}{(3 \times 2 \times 1) + (5 \times 4 \times 3 \times 2 \times 1)} \\ &= \dfrac{24 - 2}{6 + 120} \\ &= \dfrac{22}{126} \\ &= \dfrac{11}{63} \end{align*}
Question 1(iv)
Evaluate $\dfrac{(n-1)!}{(n-2)!}$.
Solution.
\begin{align*} \dfrac{(n-1)!}{(n-2)!} &= \dfrac{(n-1) \times (n-2)!}{(n-2)!} \\ &= n-1 \end{align*}
Question 1(v)
Evaluate $\dfrac{8!}{(6!)^2}$
Solution.
\begin{align*} \dfrac{8!}{(6!)^2} &= \dfrac{8 \times 7 \times 6!}{(6!) \times (6!)} \\ &= \dfrac{8 \times 7}{6!} \\ &= \dfrac{56}{720} \\ &= \dfrac{7}{90} \end{align*}
Go to