उत्तर :

(i) $8 !=1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8=40320$

(ii) $4 ! =1 \times 2 \times 3 \times 4=24$

$\quad \ 3 !=1 \times 2 \times 3=6$

$\therefore \ \ 4 ! - 3 ! =24-6=18$

उत्तर :

$3 !=1 \times 2 \times 3=6$

$4 !=1 \times 2 \times 3 \times 4=24$

$\therefore \ 3 !+4 !=6+24=30$

$7 !=1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7=5040$

$\therefore \ 3 !+4 ! \neq 7!$

उत्तर :

$\dfrac{8 !}{6 ! \times 2 !}=\dfrac{8 \times 7 \times 6 !}{6 ! \times 2 \times 1}=\dfrac{8 \times 7}{2}=28$

उत्तर :

$\dfrac{1}{6 !}+\dfrac{1}{7 !}=\dfrac{x}{8 !}$

$\Rightarrow \dfrac{1}{6 !}+\dfrac{1}{7 \times 6 !}=\dfrac{x}{8 \times 7 \times 6 !}$

$\Rightarrow \dfrac{1}{6 !}(1+\dfrac{1}{7})=\dfrac{x}{8 \times 7 \times 6 !}$

$\Rightarrow 1+\dfrac{1}{7}=\dfrac{x}{8 \times 7}$

$\Rightarrow \dfrac{8}{7}=\dfrac{x}{8 \times 7}$

$\Rightarrow x=\dfrac{8 \times 8 \times 7}{7}$

$\therefore \ x=64$

उत्तर :

(i) जब $n=6, r=2$

$ \dfrac{n !}{(n-r) !}=\dfrac{6 !}{(6-2) !}=\dfrac{6 !}{4 !}=\dfrac{6 \times 5 \times 4 !}{4 !}=30$

(ii) जब $n=9, r=5$

$ \dfrac{n !}{(n-r) !}=\dfrac{9 !}{(9-5) !}=\dfrac{9 !}{4 !}=\dfrac{9 \times 8 \times 7 \times 6 \times 5 \times 4 !}{4 !}$

$\qquad\qquad =9 \times 8 \times 7 \times 6 \times 5=15120$