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Q.
The number of three digit numbers $N = abc$ such that $a \leq b > c$, is equal to
Permutations and Combinations
Solution:
(i) $ a \neq b \neq c ($ all distinct and $c \neq 0) \Rightarrow$ Total cases $=2\left({ }^9 C _3\right)$ e.g. $1,2,3$
(ii) $a \neq b , c =0( a < b >0) \Rightarrow$ Total cases $={ }^9 C _2$
(iii) $a = b , c \neq 0( a = b > c ) \Rightarrow$ Total cases $={ }^9 C _2$
(iv) $a = b , c =0( a = b >0) \Rightarrow$ Total cases $={ }^9 C _1$
(v) $ a = c \neq 0( a < b > c ) \Rightarrow$ Total cases $={ }^9 C _2$ e.g. $1,4 \Rightarrow 141$ (formed)
$\therefore$ Number of such 3 digit numbers $=2 \cdot\left({ }^9 C _3\right)+3 \cdot\left({ }^9 C _2\right)+{ }^9 C _1=285$.
