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Combinatorial Identities and Factorials

Combinatorial Identities and Factorials

Assessment

Interactive Video

Mathematics

9th - 10th Grade

Practice Problem

Hard

Created by

Patricia Brown

FREE Resource

The video tutorial explains how to prove the identity 2nCn = 2^n * (1 * 3 * 5 * ... * (2n-1)) / n! using the combination formula. It involves expanding factorials, separating even and odd numbers, and simplifying the expression to reach the final result. The tutorial provides a step-by-step approach to understanding and solving the problem.

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10 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the main objective of the problem discussed in the video?

To solve a quadratic equation

To prove a combinatorial identity involving 2nCn

To find the roots of a polynomial

To calculate the value of pi

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Which formula is used to express combinations in the video?

NCR = n! / (n-r)!

NCR = n! * r!

NCR = n! / (r! * (n-r)!)

NCR = r! / n!

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

How is 2nCn initially expanded in terms of factorials?

n! / (2n! * n!)

2n! / (2n! * n!)

2n! / (n! * (2n-n)!)

2n! / (n! * n!)

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the purpose of separating even and odd numbers in the factorial expansion?

To simplify the expression

To make the calculation more complex

To solve a different problem

To find the value of n

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Which numbers are grouped together in the expression to simplify it?

Prime and composite numbers

Even and odd numbers

Rational and irrational numbers

Positive and negative numbers

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

How many times does the factor of 2 appear in the expression?

n/2 times

3n times

2n times

n times

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the expression rewritten as after identifying the factors of 2?

2^n * (1 * 3 * 5 * ... * (2n-1)) / n!

2^n * (1 * 2 * 3 * ... * (2n-1)) / n!

2^n * (1 * 3 * 5 * ... * n) / n!

2^n * (1 * 2 * 3 * ... * n) / n!

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