Find-the-number-of-ways-the-digits-0-1-2-and-3-can-be-permuted-to-give-rise-to-a-number-greater-than-2000-

Question Number 17770 by chux last updated on 10/Jul/17

Find the number of ways the

digits 0, 1, 2 and 3 can be permuted

to give rise to a number greater

than 2000 .

Answered by alex041103 last updated on 10/Jul/17

So we search for numbers \overset{―}{abcd} > 2000

and a \in {0, 1, 2, 3}, b \in {0, 1, 2, 3} / {a},

c \in {0, 1, 2, 3} / {a, b} and d \in {0, 1, 2, 3} / {a, b, c}

Also in order for \overset{―}{abcd} > 2000 \to a ≧ 2

So for the first digit we have 2 possabilities .

For the second, third and fourth digit

we have 3! = 6 possable permutations .

Even the smallest \overset{―}{abcd} we can have where

a ≧ 2, i . e . 2013 is bigger than 2000 . Thats

why we {don}^{'} t have restrictionsfor the last

three digits .

Total : 2 \times 3! = 2 \times 6 = 12