A quality team computes combinations of samples for an inspection lot. Given total items (n) = 7.0000; items chosen (r) = 4.0000, determine the 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

1. Step 1 — State the governing relation:

   $${}_nC_r = \dfrac{n!}{r!(n-r)!}$$

2. Step 2 — Rearrange symbolically for C:

   $$C = \dfrac{n!}{r!(n-r)!}$$

3. Step 3

   $$List the givens: total items (n) = 7.0000, items chosen (r) = 4.0000$$

4. Step 4 — Substitute the given values:

   $$C = \dfrac{7.0000!}{4.0000!(7.0000-4.0000)!}$$

5. Step 5 — Evaluate:

   $$C = 35.0000$$

6. Step 6 — Check: returning C = 35.0000 to

   $${}_nC_r = \dfrac{n!}{r!(n-r)!}$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 70.0000 — kept a factor of two that cancels in the correct rearrangement.
• 17.5000 — dropped that same factor in the other direction.
• 38.5000 — rounded an intermediate value before the final step.

Reference: FE Handbook — Permutations and Combinations

A student determines the combinations of committee members from a pool. Given total items (n) = 6.0000; items chosen (r) = 3.0000, determine the 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

1. Step 1 — State the governing relation:

   $${}_nC_r = \dfrac{n!}{r!(n-r)!}$$

2. Step 2 — Rearrange symbolically for C:

   $$C = \dfrac{n!}{r!(n-r)!}$$

3. Step 3

   $$List the givens: total items (n) = 6.0000, items chosen (r) = 3.0000$$

4. Step 4 — Substitute the given values:

   $$C = \dfrac{6.0000!}{3.0000!(6.0000-3.0000)!}$$

5. Step 5 — Evaluate:

   $$C = 20.0000$$

6. Step 6 — Check: returning C = 20.0000 to

   $${}_nC_r = \dfrac{n!}{r!(n-r)!}$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 40.0000 — kept a factor of two that cancels in the correct rearrangement.
• 10.0000 — dropped that same factor in the other direction.
• 22.0000 — rounded an intermediate value before the final step.

Reference: FE Handbook — Permutations and Combinations

The number of combinations of defective units in a batch is evaluated. Given total items (n) = 7.0000; items chosen (r) = 2.0000, determine the 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

1. Step 1 — State the governing relation:

   $${}_nC_r = \dfrac{n!}{r!(n-r)!}$$

2. Step 2 — Rearrange symbolically for C:

   $$C = \dfrac{n!}{r!(n-r)!}$$

3. Step 3

   $$List the givens: total items (n) = 7.0000, items chosen (r) = 2.0000$$

4. Step 4 — Substitute the given values:

   $$C = \dfrac{7.0000!}{2.0000!(7.0000-2.0000)!}$$

5. Step 5 — Evaluate:

   $$C = 21.0000$$

6. Step 6 — Check: returning C = 21.0000 to

   $${}_nC_r = \dfrac{n!}{r!(n-r)!}$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 42.0000 — kept a factor of two that cancels in the correct rearrangement.
• 10.5000 — dropped that same factor in the other direction.
• 23.1000 — rounded an intermediate value before the final step.

Reference: FE Handbook — Permutations and Combinations

A quality team computes combinations of samples for an inspection lot. Given total items (n) = 12.0000; items chosen (r) = 2.0000, determine the 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

1. Step 1 — State the governing relation:

   $${}_nC_r = \dfrac{n!}{r!(n-r)!}$$

2. Step 2 — Rearrange symbolically for C:

   $$C = \dfrac{n!}{r!(n-r)!}$$

3. Step 3

   $$List the givens: total items (n) = 12.0000, items chosen (r) = 2.0000$$

4. Step 4 — Substitute the given values:

   $$C = \dfrac{12.0000!}{2.0000!(12.0000-2.0000)!}$$

5. Step 5 — Evaluate:

   $$C = 66.0000$$

6. Step 6 — Check: returning C = 66.0000 to

   $${}_nC_r = \dfrac{n!}{r!(n-r)!}$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 132.0 — kept a factor of two that cancels in the correct rearrangement.
• 33.0000 — dropped that same factor in the other direction.
• 72.6000 — rounded an intermediate value before the final step.

Reference: FE Handbook — Permutations and Combinations

A student determines the combinations of committee members from a pool. Given total items (n) = 11.0000; items chosen (r) = 3.0000, determine the 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

1. Step 1 — State the governing relation:

   $${}_nC_r = \dfrac{n!}{r!(n-r)!}$$

2. Step 2 — Rearrange symbolically for C:

   $$C = \dfrac{n!}{r!(n-r)!}$$

3. Step 3

   $$List the givens: total items (n) = 11.0000, items chosen (r) = 3.0000$$

4. Step 4 — Substitute the given values:

   $$C = \dfrac{11.0000!}{3.0000!(11.0000-3.0000)!}$$

5. Step 5 — Evaluate:

   $$C = 165.0$$

6. Step 6 — Check: returning C = 165.0 to

   $${}_nC_r = \dfrac{n!}{r!(n-r)!}$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 330.0 — kept a factor of two that cancels in the correct rearrangement.
• 82.5000 — dropped that same factor in the other direction.
• 181.5 — rounded an intermediate value before the final step.

Reference: FE Handbook — Permutations and Combinations

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).

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

1. Step 1 — State the governing relation:

   $${}_nC_r = \dfrac{n!}{r!(n-r)!}$$

2. Step 2 — Rearrange symbolically for C:

   $$C = \dfrac{n!}{r!(n-r)!}$$

3. Step 3

   $$List the givens: total items (n) = 6.0000, items chosen (r) = 2.0000$$

4. Step 4 — Substitute the given values:

   $$C = \dfrac{6.0000!}{2.0000!(6.0000-2.0000)!}$$

5. Step 5 — Evaluate:

   $$C = 15.0000$$

6. 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.

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

A quality team computes combinations of samples for an inspection lot. Given total items (n) = 9.0000; items chosen (r) = 2.0000, determine the 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

1. Step 1 — State the governing relation:

   $${}_nC_r = \dfrac{n!}{r!(n-r)!}$$

2. Step 2 — Rearrange symbolically for C:

   $$C = \dfrac{n!}{r!(n-r)!}$$

3. Step 3

   $$List the givens: total items (n) = 9.0000, items chosen (r) = 2.0000$$

4. Step 4 — Substitute the given values:

   $$C = \dfrac{9.0000!}{2.0000!(9.0000-2.0000)!}$$

5. Step 5 — Evaluate:

   $$C = 36.0000$$

6. Step 6 — Check: returning C = 36.0000 to

   $${}_nC_r = \dfrac{n!}{r!(n-r)!}$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 72.0000 — kept a factor of two that cancels in the correct rearrangement.
• 18.0000 — dropped that same factor in the other direction.
• 39.6000 — rounded an intermediate value before the final step.

Reference: FE Handbook — Permutations and Combinations

A student determines the combinations of committee members from a pool. Given total items (n) = 7.0000; items chosen (r) = 5.0000, determine the 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

1. Step 1 — State the governing relation:

   $${}_nC_r = \dfrac{n!}{r!(n-r)!}$$

2. Step 2 — Rearrange symbolically for C:

   $$C = \dfrac{n!}{r!(n-r)!}$$

3. Step 3

   $$List the givens: total items (n) = 7.0000, items chosen (r) = 5.0000$$

4. Step 4 — Substitute the given values:

   $$C = \dfrac{7.0000!}{5.0000!(7.0000-5.0000)!}$$

5. Step 5 — Evaluate:

   $$C = 21.0000$$

6. Step 6 — Check: returning C = 21.0000 to

   $${}_nC_r = \dfrac{n!}{r!(n-r)!}$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 42.0000 — kept a factor of two that cancels in the correct rearrangement.
• 10.5000 — dropped that same factor in the other direction.
• 23.1000 — rounded an intermediate value before the final step.

Reference: FE Handbook — Permutations and Combinations

The number of combinations of defective units in a batch is evaluated. Given total items (n) = 11.0000; items chosen (r) = 3.0000, determine the 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

1. Step 1 — State the governing relation:

   $${}_nC_r = \dfrac{n!}{r!(n-r)!}$$

2. Step 2 — Rearrange symbolically for C:

   $$C = \dfrac{n!}{r!(n-r)!}$$

3. Step 3

   $$List the givens: total items (n) = 11.0000, items chosen (r) = 3.0000$$

4. Step 4 — Substitute the given values:

   $$C = \dfrac{11.0000!}{3.0000!(11.0000-3.0000)!}$$

5. Step 5 — Evaluate:

   $$C = 165.0$$

6. Step 6 — Check: returning C = 165.0 to

   $${}_nC_r = \dfrac{n!}{r!(n-r)!}$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 330.0 — kept a factor of two that cancels in the correct rearrangement.
• 82.5000 — dropped that same factor in the other direction.
• 181.5 — rounded an intermediate value before the final step.

Reference: FE Handbook — Permutations and Combinations

A quality team computes combinations of samples for an inspection lot. Given total items (n) = 11.0000; items chosen (r) = 1.0000, determine the 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

1. Step 1 — State the governing relation:

   $${}_nC_r = \dfrac{n!}{r!(n-r)!}$$

2. Step 2 — Rearrange symbolically for C:

   $$C = \dfrac{n!}{r!(n-r)!}$$

3. Step 3

   $$List the givens: total items (n) = 11.0000, items chosen (r) = 1.0000$$

4. Step 4 — Substitute the given values:

   $$C = \dfrac{11.0000!}{1.0000!(11.0000-1.0000)!}$$

5. Step 5 — Evaluate:

   $$C = 11.0000$$

6. Step 6 — Check: returning C = 11.0000 to

   $${}_nC_r = \dfrac{n!}{r!(n-r)!}$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 22.0000 — kept a factor of two that cancels in the correct rearrangement.
• 5.5000 — dropped that same factor in the other direction.
• 12.1000 — rounded an intermediate value before the final step.

Reference: FE Handbook — Permutations and Combinations