Chapter 11.7, Problem 6C

To determine

To calculate:The number of possibilities of choosing friends to invite to a party using permutation or combination.

Answer to Problem 6C

  $\text{10 of 30 friends can be chosen in 30045015 ways to invite to a party using combination}\text{.}$

Explanation of Solution

Given information:

Consider the information provided in the question:

  $\text{ To choose 10 of 30 friends to invite to a party using permutation or combination}\text{.}$

:

As per the question

  $\text{ To choose 10 of 30 friends to invite to a party using permutation or combination}\text{.}$

From definition,

A permutation is an arrangement of objects in which order is important.

The number of permutations of $n$ objects taken $r$ at a time is written as ${}_n{P}_r$ , where

  ${}_n{P}_r=\frac{n!}{(n-r)!}$

Whereas,

A combination is a selection of objects where the order in which the objects are chosen is not important.

The number of permutations of $n$ objects taken $r$ at a time is written as ${}_n{C}_r$ , where

  ${}_n{C}_r=\frac{{}_n{P}_r}{r!}$

Since, The order in which $\text{10 of 30 friends}$ are chosen is not important hence combination is used.

Therefore,

  ${}_{30}{C}_{10}=\frac{{}_{30}{P}_{10}}{10!}$ [Use combinations formula]

  $=\frac{30\cdot 29\cdot 28\cdot 27\cdot 26\cdot 25\cdot 24\cdot 23\cdot 22\cdot 21}{10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 2\cdot 1}$ [Write ${}_{30}{P}_{10}$ and $10!$ as products]

  $=30045015.$ [Divide out common factors and simplify]

Hence,

  $\text{10 of 30 friends can be chosen in 30045015 ways to invite to a party using combination}\text{.}$