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Q. 26

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Found in: Page 855

### Precalculus Enhanced with Graphing Utilities

Book edition 6th
Author(s) Sullivan
Pages 1200 pages
ISBN 9780321795465

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

The list of ordered arrangements 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.

The value of the given permutation is 120.

See the step by step solution

## Step 1. Given information.

3 objects are chosen from 6 objects 1, 2, 3, 4, 5, and 6 at a time without repetition.

The given permutation is $P\left(6,3\right)$.

## Step 2. Determine the list of ordered arrangements.

The list of ordered arrangements 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.

## Step 3. Determine the value of permutation.

We know that,

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

Using the above formula, we get

$\begin{array}{rcl}P\left(6,3\right)& =& \frac{6!}{\left(6-3\right)!}\\ & =& \frac{6!}{3!}\\ & =& \frac{6×5×4×3!}{3!}\\ & =& 120\end{array}$