Question

What is the average of all possible five-digit numbers that can be formed by using each of the digits 2, 1, 8, 7, and 6 exactly once?​

Answer 1

Solution :-

Since each digit can be in every position and there are no repetition, so is a permutation without repetition. Therefore, Total possible numbers are = 5! = 5*4*3*2*1 = 120

Now, we have to find the sum of all these numbers and divide by 120 to get the average.

Now the numbers will be as :- 21876, 21867, 28761, 28167, 27861, 28716, 26781 ________and so on

Now if you notice 2 will come at 10000 place 24 times and similarly for 5,6,7,8 as they will also come at 10000 place 24 times . ( Each = 120/5 = 24 Times. )

Again, 1 will come at 1000 place 24 times and same for other digits.

And so on for 100th, 10th and units place..

Therefore,

→ The sum of these numbers will be :- [(2+1+8+7+6)*10000 + (2+1+8+7+6)*1000 + (2+1+8+7+6)*100 + (2+1+8+7+6)*10 + (2+1+8+7+6)*1] * 24

→ (24*10000 + 24*1000 + 24*100 + 24*10 + 24*1)*24

→ (240000 + 24000 + 2400 + 240 + 24) * 24

→ (266,664) * 24

→ 6,399,936

Hence,

→ Average of These 120 Numbers = (Total sum) / (Total No.) = (6399936) / 120 = 53,332.8 (Ans.)