Chapter 11.7, Problem 2C

To determine

To calculate:The number of combinations.

Answer to Problem 2C

  ${}_6{C}_3\text{ is equal to 20}\text{.}$

Explanation of Solution

Given information:

Consider the information provided in the question:

  ${}_6{C}_3$

Formula used:

To find the number of combinations of $n$ objects taken $r$ at a time, divide the number of permutations of $n$ objects taken $r$ at a time by $r!$

  ${}_n{C}_r=\frac{{}_n{P}_r}{r!}$

Calculation:

As per the problem,

  ${}_6{C}_3$

To calculate the number of combinations, use formula

  ${}_n{C}_r=\frac{{}_n{P}_r}{r!}$

Substitute $\text{6 for }n\text{ and 3 for }r$

  ${}_6{C}_3=\frac{{}_6{P}_3}{3!}$

  $=\frac{6\cdot 5\cdot 4\cdot 3}{3\cdot 2\cdot 1}$ [Write ${}_6{P}_3$ and $3!$ as products]

  $=20$ [Divide out common factors and simplify]

Hence,

  ${}_6{C}_3\text{ is equal to 20}\text{.}$