Question

What is the average of all possible five-digit numbers that can be formed
by using each of the digits 6, 7,5,9, and 2 exactly once?​

Answer 1

Answer:

64,443.80

Step-by-step explanation:

First of all, here repetition of digits is not allowed.

So, total 5-digits number formed by 6,7,5,9 & 2 will be :- 5*4*3*2*1 = 120

(You can solve total 5 digits number by permutation method :-⁵ P ₅ = 5!/(5-5)! = 5!/0! = 5*4*3*2*1 = 120)  Note:- Value of 0! = 1.

Now, For finding sum of these 120 numbers :

We know that all the digits will come in all 5 positions.

For example :- 25796, 52796, 65972, etc.. It means all the digits will surely come to all the positions (unit place, tens place, hundreds place, thousand place & ten thousand place).

Now, The main question is how many times each digit will come to all position ??

so, it will come total 120 ÷ 5 = 24 times to all the places.

So, sum of all the digits = 24 (6+5+7+9+2) = 24*29 = 696

Now, total sum of all numbers formed by these 5 digits = 696* (10⁴ + 10³ + 10² + 10 + 1) = 696*11,111 = 77,33,256.

Now, Average of all 120 numbers formed = Sum of all numbers formed / 120

                                                                      =  77,33,256 / 120

                                                                      = 64,443.80

Hope you understood the concept !!!