Chapter 6 Permutations And Combinations EXERCISE 6.2
EXERCISE 6.2
1. Evaluate
(i) 8 !
(ii) 4 ! -3 !
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Answer :
(i) $8 !=1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8=40320$
(ii) 4 ! $=1 \times 2 \times 3 \times 4=24$
$3 !=1 \times 2 \times 3=6$
$\therefore 4$ ! - 3 ! $=24-6=18$
2. Is $3 !+4 !=7 !$ ?
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Answer :
$3 !=1 \times 2 \times 3=6$
$4 !=1 \times 2 \times 3 \times 4=24$
$\therefore 3 !+4 !=6+24=30$
$7 !=1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7=5040$
$\therefore 3 !+4 ! \neq 7$ !
3. Compute $\dfrac{8 !}{6 ! \times 2 !}$
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Answer :
$\dfrac{8 !}{6 ! \times 2 !}=\dfrac{8 \times 7 \times 6 !}{6 ! \times 2 \times 1}=\dfrac{8 \times 7}{2}=28$
4. If $\dfrac{1}{6 !}+\dfrac{1}{7 !}=\dfrac{x}{8 !}$, find $x$
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Answer :
$\dfrac{1}{6 !}+\dfrac{1}{7 !}=\dfrac{x}{8 !}$
$\Rightarrow \dfrac{1}{6 !}+\dfrac{1}{7 \times 6 !}=\dfrac{x}{8 \times 7 \times 6 !}$
$\Rightarrow \dfrac{1}{6 !}(1+\dfrac{1}{7})=\dfrac{x}{8 \times 7 \times 6 !}$
$\Rightarrow 1+\dfrac{1}{7}=\dfrac{x}{8 \times 7}$
$\Rightarrow \dfrac{8}{7}=\dfrac{x}{8 \times 7}$
$\Rightarrow x=\dfrac{8 \times 8 \times 7}{7}$
$\therefore x=64$
5. Evaluate $\dfrac{n !}{(n-r) !}$, when
(i) $n=6, r=2$
(ii) $n=9, r=5$.
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Answer :
(i) When $n=6, r=2, \dfrac{n !}{(n-r) !}=\dfrac{6 !}{(6-2) !}=\dfrac{6 !}{4 !}=\dfrac{6 \times 5 \times 4 !}{4 !}=30$
(ii) When $n=9, r=5, \dfrac{n !}{(n-r) !}=\dfrac{9 !}{(9-5) !}=\dfrac{9 !}{4 !}=\dfrac{9 \times 8 \times 7 \times 6 \times 5 \times 4 !}{4 !}$
$=9 \times 8 \times 7 \times 6 \times 5=15120$
