Skip to content
Back to exploration
Discrete mathematics / English

COMBINATIONS with REPETITION - DISCRETE MATHEMATICS

TrevTutor · YouTube · 13:34

Open original
READ & KEEP

The explanation, unpacked.

Reviewed learning material · Video analysis · English
Read the full overview

Connect combinations with repetition to stars and bars, then account for a variable's lower bound.

Use the learning inspector for key ideas and moments, or open the reading tabs for the complete notes.

Chapters

1:30Selections with repetition6:00Stars and bars12:00Shift a lower bound

Learning script

Generated from the video's visuals and explanation; not verbatim speech.

Choosing r objects from n types, with repetition allowed and order ignored, gives (n+r−1r)\binom{n+r-1}{r} possibilities. For four types and three objects this is (63)=20\binom{6}{3}=20.

Nonnegative solutions of x1+⋯+xn=rx_1+\cdots+x_n=r correspond to r identical stars separated into n groups by n−1n-1 bars. Empty groups are allowed.

For x1+x2+x3=9x_1+x_2+x_3=9 with x3≥2x_3\ge2 and the others nonnegative, substitute x3′=x3−2x_3'=x_3-2. The remaining total is 7, giving (92)=36\binom{9}{2}=36 solutions.

Knowledge cards

01

Selections with repetition

Choosing r objects from n types, with repetition allowed and order ignored, gives (n+r−1r)\binom{n+r-1}{r} possibilities. For four types and three objects this is (63)=20\binom{6}{3}=20.

02

Stars and bars

Nonnegative solutions of x1+⋯+xn=rx_1+\cdots+x_n=r correspond to r identical stars separated into n groups by n−1n-1 bars. Empty groups are allowed.

03

Shift a lower bound

For x1+x2+x3=9x_1+x_2+x_3=9 with x3≥2x_3\ge2 and the others nonnegative, substitute x3′=x3−2x_3'=x_3-2. The remaining total is 7, giving (92)=36\binom{9}{2}=36 solutions.

Explore the knowledge in this video

Open video knowledge graph →

  • Combinations ExplanationAt 1:30
    Why this connection?

    Choosing r objects from n types, with repetition allowed and order ignored, gives (n+r−1r)\binom{n+r-1}{r} possibilities. For four types and three objects this is (63)=20\binom{6}{3}=20. Nonnegative solutions of x1+⋯+xn=rx_1+\cdots+x_n=r correspond to r identical stars separated into n groups by n−1n-1 bars. Empty groups are allowed. For x1+x2+x3=9x_1+x_2+x_3=9 with x3≥2x_3\ge2 and the others nonnegative, substitute x3′=x3−2x_3'=x_3-2. The remaining total is 7, giving (92)=36\binom{9}{2}=36 solutions.

Questions this video answers

Find a method

↗
Find a method

↗
Find a method

↗