Chapter 13.2, Problem 26AYU

List all the ordered arrangements of 6 objects 1, 2, 3, 4, 5, and 6 choosing 3 at a time without repetition. What is $P\left(6,3\right)$?

To determine

To write:

a. All the ordered arrangements of 6 objects 1, 2, 3, 4, 5 and 6 choosing 3 at a time without repetition.

b. What is $P\left(6, 3\right)?$

Answer to Problem 26AYU

a. $\{123,124,125,126,132,134,135,136,142,143,145,146,152,153,154,156,162,163,164,165, 213,214,215,216,231,234,235,236,241,243,245,246, 251,253,254,256,261,263,264,265,312,314,315,316,321,324,325,326,341,342,345,346,351,352,354,356,361,362,364,365,412,413,415,416,421,423,425,426,431,432,435,436,451,452,453,456,461,462,463,465,512,513,514,516,521,523,524,526,531,532,534,536,541,542,543,546,561,562,563,564,612,613,614,615,621,623,624,625,631,632,634,635,641,642,643,645,651,652,653,654\}$

b. 120

Explanation of Solution

Given:

The 6 objects 1, 2, 3, 4, 5 and 6 number of objects to be chosen is 3 without repetition.

Formula used:

$P\left(n,r\right)=\frac{n!}{\left(n-r\right)!}$

Calculation:

The list of ordered arrangements of 6 objects 1, 2, 3, 4, 5 and 6 choosing 3 at a time without repetition.

Arrangement starting with the object 1 is $\{123,124,125,126,132,134,135,136,142,143,145,146,152,153,154,156,162,163,164,165\}$.

Arrangement starting with the object 2 is $\{213,214,215,216,231,234,235,236,241,243,245,246, 251,253,254,256,261,263,264,265\}$.

Arrangement starting with the object 3 is $\left\{312,314,315,316,321,324,325,326,341,342,345,346,351,352,354,356,361,362,364,365\right\}$.

Arrangement starting with the object 4 is $\left\{412,413,415,416,421,423,425,426,431,432,435,436,451,452,453,456,461,462,463,465\right\}$.

Arrangement starting with the object 5 is $\left\{512,513,514,516,521,523,524,526,531,532,534,536,541,542,543,546,561,562,563,564\right\}$

Arrangement starting with the object 6 is $\left\{612,613,614,615,621,623,624,625,631,632,634,635,641,642,643,645,651,652,653,654\right\}$

The list of ordered arrangements of 6 objects 1, 2, 3, 4, 5 and 6 choosing 3 at a time without repetition is $\{123,124,125,126,132,134,135,136,142,143,145,146,152,153,154,156,162,163,164,165, 213,214,215,216,231,234,235,236,241,243,245,246, 251,253,254,256,261,263,264,265,312,314,315,316,321,324,325,326,341,342,345,346,351,352,354,356,361,362,364,365,412,413,415,416,421,423,425,426,431,432,435,436,451,452,453,456,461,462,463,465,512,513,514,516,521,523,524,526,531,532,534,536,541,542,543,546,561,562,563,564,612,613,614,615,621,623,624,625,631,632,634,635,641,642,643,645,651,652,653,654\}$

$P\left(6,3\right)=\frac{6!}{\left(4-3\right)!}=\frac{6\text{*}5\text{*}4\text{*}3\text{*}2\text{*}1}{3\text{*}2\text{*}1}=120$