A probability problem by RAJ RAJPUT
Probability
Level
3
if all the six digits numbers 1,2,3,4,5,6 are arranged in increasing order , find the 100th number,such that last digit should not be 5 in any number ....
The answer is 213654.
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1 solution
Let the 6 digits be represented as ABCDEF.
To find the increasing order of numbers, we first put A =1. Since F cannot be equal to 5, the number of ways we can decide on digits BCDEF is 4 * 3 *2 * 1 *4 ( note the multiplicative order) i.e. 96 numbers.
Now the next number should be 213456. Also note that the next larger number 213465 is invalid given the condition.
Therefore, we rearrange the last three digits instead of the last two ( i.e. we rearranged DEF). The digits to be used for rearrangement are 4,5 and 6. The number of ways to do this is 2 * 1 * 2 in that order.
Now we can simply write down the numbers.
213546 - 98th number
213564 - 99th number
213654 - 100 th number ( we omit 213645)