First slide

Permutations

Question

$T h e s u m o f a l l t h e f o u r d i g i t d i s t i n c t n u m b e r s t h a t c a n b e f o r m e d w i t h t h e d i g i t s 1, 2, 2, a n d 3 i s$

Moderate

A

$22264$
B

$122664$
C

$26664$
D

122234

Solution

$\begin{array}{l} 1 at unit place \Rightarrow number of such numbers is \frac{3!}{2!} = 3 \\ 2 at unit place \Rightarrow number of such numbers is 3! = 6 \\ 3 at unit place \Rightarrow number of such numbers is \frac{3!}{2!} = 3 \\ Hence the sum of the digits in the unit place is 3 \times 1 + 6 \times 2 + 3 \times 3 = 24 \\ The sum of all 4 digit numbers is 1111 \times 24 = 26664 \end{array}$

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