Permutations and Combinations

Probability and Statistics · FE Reference Handbook section

Probability and Statistics

7 formulas

10 exam-style examples

~59 min

All Probability and Statistics lectures

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

A permutation is a particular sequence of a given set of objects. A combination is the set itself without reference to order.
1. The number of different permutations of n distinct objects taken r at a time is
nPr is an alternative notation for P(n,r)
2. The number of different combinations of n distinct objects taken r at a time is
nCr and e o are alternative notations for C(n,r)

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Permutations — solve for number of permutations — Permutations and Combinations

An engineer counts the permutations of test sequences for a QA plan. Given total items (n) = 6.0000; items chosen (r) = 3.0000, determine the number of permutations (P).

Given

$total items (n) = 6.0000$
$items chosen (r) = 3.0000$

Find

number of permutations (P)

Start with the thinking

The governing relation printed in this handbook section is Permutations.
Everything except P is given, so isolate P symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Permutations and combinations count the ordered arrangements of r items chosen from n distinct items.

Step-by-step solution

Step 1 — State the governing relation:
${}_nP_r = \dfrac{n!}{(n-r)!}$
Step 2 — Rearrange symbolically for P:
$P = \dfrac{n!}{(n-r)!}$
Step 3
$List the givens: total items (n) = 6.0000, items chosen (r) = 3.0000$
Step 4 — Substitute the given values:
$P = \dfrac{6.0000!}{(6.0000-3.0000)!}$
Step 5 — Evaluate:
$P = 120.0$
Step 6 — Check: returning P = 120.0 to
${}_nP_r = \dfrac{n!}{(n-r)!}$
reproduces the given quantities, and both sides carry the same units.

Answer:

P = 120.0

Why the other options are there

240.0 — kept a factor of two that cancels in the correct rearrangement.
60.0000 — dropped that same factor in the other direction.
132.0 — rounded an intermediate value before the final step.

Reference: FE Handbook — Permutations and Combinations

Example 2

Combinations — solve for number of combinations — Permutations and Combinations (2)

A student determines the combinations of committee members from a pool. Given total items (n) = 9.0000; items chosen (r) = 3.0000, determine the number of combinations (C).

Given

$total items (n) = 9.0000$
$items chosen (r) = 3.0000$

Find

number of combinations (C)

Start with the thinking

The governing relation printed in this handbook section is Combinations.
Everything except C is given, so isolate C symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Combinations count the unordered subsets of r items chosen from n distinct items.

Step-by-step solution

Step 1 — State the governing relation:
${}_nC_r = \dfrac{n!}{r!(n-r)!}$
Step 2 — Rearrange symbolically for C:
$C = \dfrac{n!}{r!(n-r)!}$
Step 3
$List the givens: total items (n) = 9.0000, items chosen (r) = 3.0000$
Step 4 — Substitute the given values:
$C = \dfrac{9.0000!}{3.0000!(9.0000-3.0000)!}$
Step 5 — Evaluate:
$C = 84.0000$
Step 6 — Check: returning C = 84.0000 to
${}_nC_r = \dfrac{n!}{r!(n-r)!}$
reproduces the given quantities, and both sides carry the same units.

Answer:

C = 84.0000

Why the other options are there

168.0 — kept a factor of two that cancels in the correct rearrangement.
42.0000 — dropped that same factor in the other direction.
92.4000 — rounded an intermediate value before the final step.

Reference: FE Handbook — Permutations and Combinations

Example 3

Permutations — solve for number of permutations (case 2) — Permutations and Combinations (3)

A student computes permutations and combinations for a scheduling problem. Given total items (n) = 8.0000; items chosen (r) = 1.0000, determine the number of permutations (P).

Given

$total items (n) = 8.0000$
$items chosen (r) = 1.0000$

Find

number of permutations (P)

Start with the thinking

The governing relation printed in this handbook section is Permutations.
Everything except P is given, so isolate P symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Permutations and combinations count the ordered arrangements of r items chosen from n distinct items.

Step-by-step solution

Step 1 — State the governing relation:
${}_nP_r = \dfrac{n!}{(n-r)!}$
Step 2 — Rearrange symbolically for P:
$P = \dfrac{n!}{(n-r)!}$
Step 3
$List the givens: total items (n) = 8.0000, items chosen (r) = 1.0000$
Step 4 — Substitute the given values:
$P = \dfrac{8.0000!}{(8.0000-1.0000)!}$
Step 5 — Evaluate:
$P = 8.0000$
Step 6 — Check: returning P = 8.0000 to
${}_nP_r = \dfrac{n!}{(n-r)!}$
reproduces the given quantities, and both sides carry the same units.

Answer:

P = 8.0000

Why the other options are there

16.0000 — kept a factor of two that cancels in the correct rearrangement.
4.0000 — dropped that same factor in the other direction.
8.8000 — rounded an intermediate value before the final step.

Reference: FE Handbook — Permutations and Combinations

Example 4

Combinations — solve for number of combinations (case 2) — Permutations and Combinations (4)

The number of combinations of defective units in a batch is evaluated. Given total items (n) = 5.0000; items chosen (r) = 4.0000, determine the number of combinations (C).

Given

$total items (n) = 5.0000$
$items chosen (r) = 4.0000$

Find

number of combinations (C)

Start with the thinking

The governing relation printed in this handbook section is Combinations.
Everything except C is given, so isolate C symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Combinations count the unordered subsets of r items chosen from n distinct items.

Step-by-step solution

Step 1 — State the governing relation:
${}_nC_r = \dfrac{n!}{r!(n-r)!}$
Step 2 — Rearrange symbolically for C:
$C = \dfrac{n!}{r!(n-r)!}$
Step 3
$List the givens: total items (n) = 5.0000, items chosen (r) = 4.0000$
Step 4 — Substitute the given values:
$C = \dfrac{5.0000!}{4.0000!(5.0000-4.0000)!}$
Step 5 — Evaluate:
$C = 5.0000$
Step 6 — Check: returning C = 5.0000 to
${}_nC_r = \dfrac{n!}{r!(n-r)!}$
reproduces the given quantities, and both sides carry the same units.

Answer:

C = 5.0000

Why the other options are there

10.0000 — kept a factor of two that cancels in the correct rearrangement.
2.5000 — dropped that same factor in the other direction.
5.5000 — rounded an intermediate value before the final step.

Reference: FE Handbook — Permutations and Combinations

Example 5

Permutations — solve for number of permutations (case 3) — Permutations and Combinations (5)

The number of permutations of finish orders in a lab trial is calculated. Given total items (n) = 7.0000; items chosen (r) = 4.0000, determine the number of permutations (P).

Given

$total items (n) = 7.0000$
$items chosen (r) = 4.0000$

Find

number of permutations (P)

Start with the thinking

The governing relation printed in this handbook section is Permutations.
Everything except P is given, so isolate P symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Permutations and combinations count the ordered arrangements of r items chosen from n distinct items.

Step-by-step solution

Step 1 — State the governing relation:
${}_nP_r = \dfrac{n!}{(n-r)!}$
Step 2 — Rearrange symbolically for P:
$P = \dfrac{n!}{(n-r)!}$
Step 3
$List the givens: total items (n) = 7.0000, items chosen (r) = 4.0000$
Step 4 — Substitute the given values:
$P = \dfrac{7.0000!}{(7.0000-4.0000)!}$
Step 5 — Evaluate:
$P = 840.0$
Step 6 — Check: returning P = 840.0 to
${}_nP_r = \dfrac{n!}{(n-r)!}$
reproduces the given quantities, and both sides carry the same units.

Answer:

P = 840.0

Why the other options are there

1,680 — kept a factor of two that cancels in the correct rearrangement.
420.0 — dropped that same factor in the other direction.
924.0 — rounded an intermediate value before the final step.

Reference: FE Handbook — Permutations and Combinations

Example 6

Combinations — solve for number of combinations (case 3) — Permutations and Combinations (6)

A quality team computes combinations of samples for an inspection lot. Given total items (n) = 6.0000; items chosen (r) = 5.0000, determine the number of combinations (C).

Given

$total items (n) = 6.0000$
$items chosen (r) = 5.0000$

Find

number of combinations (C)

Start with the thinking

The governing relation printed in this handbook section is Combinations.
Everything except C is given, so isolate C symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Combinations count the unordered subsets of r items chosen from n distinct items.

Step-by-step solution

Step 1 — State the governing relation:
${}_nC_r = \dfrac{n!}{r!(n-r)!}$
Step 2 — Rearrange symbolically for C:
$C = \dfrac{n!}{r!(n-r)!}$
Step 3
$List the givens: total items (n) = 6.0000, items chosen (r) = 5.0000$
Step 4 — Substitute the given values:
$C = \dfrac{6.0000!}{5.0000!(6.0000-5.0000)!}$
Step 5 — Evaluate:
$C = 6.0000$
Step 6 — Check: returning C = 6.0000 to
${}_nC_r = \dfrac{n!}{r!(n-r)!}$
reproduces the given quantities, and both sides carry the same units.

Answer:

C = 6.0000

Why the other options are there

12.0000 — kept a factor of two that cancels in the correct rearrangement.
3.0000 — dropped that same factor in the other direction.
6.6000 — rounded an intermediate value before the final step.

Reference: FE Handbook — Permutations and Combinations

Example 7

Permutations — solve for number of permutations (case 4) — Permutations and Combinations (7)

An engineer counts the permutations of test sequences for a QA plan. Given total items (n) = 7.0000; items chosen (r) = 4.0000, determine the number of permutations (P).

Given

$total items (n) = 7.0000$
$items chosen (r) = 4.0000$

Find

number of permutations (P)

Start with the thinking

The governing relation printed in this handbook section is Permutations.
Everything except P is given, so isolate P symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Permutations and combinations count the ordered arrangements of r items chosen from n distinct items.

Step-by-step solution

Step 1 — State the governing relation:
${}_nP_r = \dfrac{n!}{(n-r)!}$
Step 2 — Rearrange symbolically for P:
$P = \dfrac{n!}{(n-r)!}$
Step 3
$List the givens: total items (n) = 7.0000, items chosen (r) = 4.0000$
Step 4 — Substitute the given values:
$P = \dfrac{7.0000!}{(7.0000-4.0000)!}$
Step 5 — Evaluate:
$P = 840.0$
Step 6 — Check: returning P = 840.0 to
${}_nP_r = \dfrac{n!}{(n-r)!}$
reproduces the given quantities, and both sides carry the same units.

Answer:

P = 840.0

Why the other options are there

1,680 — kept a factor of two that cancels in the correct rearrangement.
420.0 — dropped that same factor in the other direction.
924.0 — rounded an intermediate value before the final step.

Reference: FE Handbook — Permutations and Combinations

Example 8

Combinations — solve for number of combinations (case 4) — Permutations and Combinations (8)

A student determines the combinations of committee members from a pool. Given total items (n) = 9.0000; items chosen (r) = 4.0000, determine the number of combinations (C).

Given

$total items (n) = 9.0000$
$items chosen (r) = 4.0000$

Find

number of combinations (C)

Start with the thinking

The governing relation printed in this handbook section is Combinations.
Everything except C is given, so isolate C symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Combinations count the unordered subsets of r items chosen from n distinct items.

Step-by-step solution

Step 1 — State the governing relation:
${}_nC_r = \dfrac{n!}{r!(n-r)!}$
Step 2 — Rearrange symbolically for C:
$C = \dfrac{n!}{r!(n-r)!}$
Step 3
$List the givens: total items (n) = 9.0000, items chosen (r) = 4.0000$
Step 4 — Substitute the given values:
$C = \dfrac{9.0000!}{4.0000!(9.0000-4.0000)!}$
Step 5 — Evaluate:
$C = 126.0$
Step 6 — Check: returning C = 126.0 to
${}_nC_r = \dfrac{n!}{r!(n-r)!}$
reproduces the given quantities, and both sides carry the same units.

Answer:

C = 126.0

Why the other options are there

252.0 — kept a factor of two that cancels in the correct rearrangement.
63.0000 — dropped that same factor in the other direction.
138.6 — rounded an intermediate value before the final step.

Reference: FE Handbook — Permutations and Combinations

Example 9

Permutations — solve for number of permutations (case 5) — Permutations and Combinations (9)

A student computes permutations and combinations for a scheduling problem. Given total items (n) = 6.0000; items chosen (r) = 1.0000, determine the number of permutations (P).

Given

$total items (n) = 6.0000$
$items chosen (r) = 1.0000$

Find

number of permutations (P)

Start with the thinking

The governing relation printed in this handbook section is Permutations.
Everything except P is given, so isolate P symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Permutations and combinations count the ordered arrangements of r items chosen from n distinct items.

Step-by-step solution

Step 1 — State the governing relation:
${}_nP_r = \dfrac{n!}{(n-r)!}$
Step 2 — Rearrange symbolically for P:
$P = \dfrac{n!}{(n-r)!}$
Step 3
$List the givens: total items (n) = 6.0000, items chosen (r) = 1.0000$
Step 4 — Substitute the given values:
$P = \dfrac{6.0000!}{(6.0000-1.0000)!}$
Step 5 — Evaluate:
$P = 6.0000$
Step 6 — Check: returning P = 6.0000 to
${}_nP_r = \dfrac{n!}{(n-r)!}$
reproduces the given quantities, and both sides carry the same units.

Answer:

P = 6.0000

Why the other options are there

12.0000 — kept a factor of two that cancels in the correct rearrangement.
3.0000 — dropped that same factor in the other direction.
6.6000 — rounded an intermediate value before the final step.

Reference: FE Handbook — Permutations and Combinations

Example 10

Combinations — solve for number of combinations (case 5) — Permutations and Combinations (10)

The number of combinations of defective units in a batch is evaluated. Given total items (n) = 6.0000; items chosen (r) = 2.0000, determine the number of combinations (C).

Given

$total items (n) = 6.0000$
$items chosen (r) = 2.0000$

Find

number of combinations (C)

Start with the thinking

The governing relation printed in this handbook section is Combinations.
Everything except C is given, so isolate C symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Combinations count the unordered subsets of r items chosen from n distinct items.

Step-by-step solution

Step 1 — State the governing relation:
${}_nC_r = \dfrac{n!}{r!(n-r)!}$
Step 2 — Rearrange symbolically for C:
$C = \dfrac{n!}{r!(n-r)!}$
Step 3
$List the givens: total items (n) = 6.0000, items chosen (r) = 2.0000$
Step 4 — Substitute the given values:
$C = \dfrac{6.0000!}{2.0000!(6.0000-2.0000)!}$
Step 5 — Evaluate:
$C = 15.0000$
Step 6 — Check: returning C = 15.0000 to
${}_nC_r = \dfrac{n!}{r!(n-r)!}$
reproduces the given quantities, and both sides carry the same units.

Answer:

C = 15.0000

Why the other options are there

30.0000 — kept a factor of two that cancels in the correct rearrangement.
7.5000 — dropped that same factor in the other direction.
16.5000 — rounded an intermediate value before the final step.

Reference: FE Handbook — Permutations and Combinations