For a science exhibition, a student coordinator needs to choose 4 project displays from a … — JAMB UTME

For a science exhibition, a student coordinator needs to choose $4$ project displays from a pool of $6$ chemistry projects and $4$ physics projects. If the selection must include at least $2$ chemistry projects, in how many distinct ways can the projects be selected?

A
$170$
$185$ Answer
C
$210$
D
$1,200$

Answer

The total number of distinct ways to select the projects is

185

The correct total number of ways is

185

. The condition requiring at least

2

chemistry projects splits the total selection into three mutually exclusive scenarios: selecting

2

chemistry and

2

physics projects (

90

ways),

3

chemistry and

1

physics project (

80

ways), or

4

chemistry projects (

15

ways). Adding these together yields

90 + 80 + 15 = 185

Step-by-Step Solution

Identify the distinct valid cases for selecting

4

projects with at least

2

chemistry projects.

Case 1:

2

chemistry and

2

physics; Case 2:

3

chemistry and

1

physics; Case 3:

4

chemistry and

0

physics.

The phrase 'at least 2 chemistry projects' means the number of chemistry projects can be

2

3

, or

4

Calculate the combinations for Case 1 (

2

chemistry,

2

physics).

$\binom{6}{2} \times \binom{4}{2} = 15 \times 6 = 90$ ways.

We select

2

projects out of

6

chemistry projects and

2

out of

4

physics projects using the combination formula $\binom{n}{r} = \frac{n!}{r!(n-r)!}$.

Calculate the combinations for Case 2 (

3

chemistry,

1

physics).

$\binom{6}{3} \times \binom{4}{1} = 20 \times 4 = 80$ ways.

We select

3

projects out of

6

chemistry projects and

1

out of

4

physics projects.

Calculate the combinations for Case 3 (

4

chemistry,

0

physics).

$\binom{6}{4} \times \binom{4}{0} = 15 \times 1 = 15$ ways.

We select

4

projects out of

6

chemistry projects and

0

out of

4

physics projects.

Sum the number of ways from all mutually exclusive cases.

$90 + 80 + 15 = 185$ ways.

By the addition principle of counting, the total number of ways is the sum of the ways from each independent case.

Key Concept

Combinations with Constraints (At Least / At Most)

Question

Answer

Step-by-Step Solution

Key Concept

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