Counting & Binomials | IB Math AAHL | Perplex

Counting & Binomials

Video Reviews

Watch comprehensive video reviews for Counting & Binomials, designed for final exam preparation. Each video includes integrated problems you can solve alongside detailed solutions.

Not your average video:

Interactive Problems: Solve problems alongside the video with step-by-step guidance and detailed solutions.

Exam Preparation: Complete unit reviews designed for final exam preparation with all key concepts covered systematically.

Expert Teaching: High-quality instruction from Perplex co-founder James Mullen with clear explanations, worked examples, and exam tips.

Not your average video:

Interactive Problems: Solve problems alongside the video with step-by-step guidance and detailed solutions.

Exam Preparation: Complete unit reviews designed for final exam preparation with all key concepts covered systematically.

Expert Teaching: High-quality instruction from Perplex co-founder James Mullen with clear explanations, worked examples, and exam tips.

SL TheoryBinomial ProblemsCounting (HL)

The video will automatically pause when it reaches a problem.

Orderings where certain items need to stay apart

AHL AA 1.10

How many ways can the items B1,B2,B3,B4,B5 be ordered under the restriction that B1 and B2 must not be together.

To solve questions like these, we recognize that for B1 and B2 not to be together, there must be at least one item between them. To guarantee this, we first arrange the other items:

3!=6

orderings for B3,B4 and B5.

For argument's sake, consider the arrangement B3B4B5. Before, after and between each of these we put a slot:

B3B3B3B4B3B5B3

There are then 4 slots for B1 and B2, and they cannot be placed in the same slot. This gives 4⋅3=12 arrangements for B1 and B2 after arranging the other items. In total, there are

12⋅6=72

arrangements.

Orderings where certain items need to stay apart

AHL AA 1.10

How many ways can the items B1,B2,B3,B4,B5 be ordered under the restriction that B1 and B2 must not be together.

To solve questions like these, we recognize that for B1 and B2 not to be together, there must be at least one item between them. To guarantee this, we first arrange the other items:

3!=6

orderings for B3,B4 and B5.

For argument's sake, consider the arrangement B3B4B5. Before, after and between each of these we put a slot:

B3B3B3B4B3B5B3

There are then 4 slots for B1 and B2, and they cannot be placed in the same slot. This gives 4⋅3=12 arrangements for B1 and B2 after arranging the other items. In total, there are

12⋅6=72

arrangements.