Chapter 13.1, Problem 53E

a.

To determine

To calculate: The average of six permutations of 2, 5 and 9.

Answer to Problem 53E

The average of six permutations of 2, 5 and 9is $592$ .

Explanation of Solution

Given information:

The average of permutations formed by 3 digits say a, b and c is equivalent to $37\left(a+b+c\right)$ .

Calculation:

Consider the provided information that permutation of 2, 5 and 9 are to be computed.

The permutations are 259, 295, 529, 592, 925 and 952.

The average of the numbers obtained above is,

  $\begin{array}{c}\frac{259+295+529+592+925+952}{6}=\frac{3552}{6}\\ =592\end{array}$

The average of permutations formed by 3 digits say a, b and c is equivalent to $37\left(a+b+c\right)$ .

Now, $37\left(2+5+9\right)=592$

Therefore, the average of six permutations formed by digits 2, 5 and 9 is equal to $37\left(2+5+9\right)=592$ .

Thus, the average of six permutations of 2, 5 and 9 is $592$ .

b.

To determine

To explain: The average of six permutations of any three digits is equal to 37 times the sum of the 3 digits is true or not.

Answer to Problem 53E

The average of six permutations of any three digits is equal to 37 times the sum of the 3 digits is true.

Explanation of Solution

Given information:

The arbitrary 3 digits.

Calculation:

Consider the provided information that arbitrary 3 digits are provided say $a,b\text{ and }c$ .

The permutations of the above digits are abc, acb, bac, bca, cabandcba .

The average of the numbers obtained above is,

  $\begin{array}{c}\frac{abc+acb+bac+bca+cab+cba}{6}=\frac{222\left(a+b+c\right)}{6}\\ =37\left(a+b+c\right)\end{array}$

The average of six permutations of any three digits is equal to 37 times the sum of the 3 digits is true.