An urn contains 5 red marbles, 4 black marbles and 3 white marbles. Then the number of ways in which 4 marbles can be drawn so that at the most three of them are red is ___________.
Answer (integer)
490
Solution
Here 5 red marbels and 7 non red marbels presents.
<br><br>No of ways 4 marbels can be chosen where atmost 3 red marbels can be present.
<br><br><b>Case 1:</b> When 3 red marbels present
<br><br>No of ways = <sup>5</sup>C<sub>3</sub> $\times$ <sup>7</sup>C<sub>1</sub>
<br><br><b>Case 2:</b> When 2 red marbels present
<br><br>No of ways = <sup>5</sup>C<sub>2</sub> $\times$ <sup>7</sup>C<sub>2</sub>
<br><br><b>Case 3:</b> When 1 red marbels present
<br><br>No of ways = <sup>5</sup>C<sub>1</sub> $\times$ <sup>7</sup>C<sub>3</sub>
<br><br><b>Case 4:</b> When 0 red marbels present
<br><br>No of ways = <sup>5</sup>C<sub>0</sub> $\times$ <sup>7</sup>C<sub>4</sub>
<br><br>$\therefore$ Total number of ways
<br><br> = <sup>5</sup>C<sub>3</sub> $\times$ <sup>7</sup>C<sub>1</sub> + <sup>5</sup>C<sub>2</sub> $\times$ <sup>7</sup>C<sub>2</sub> + <sup>5</sup>C<sub>1</sub> $\times$ <sup>7</sup>C<sub>3</sub> + <sup>5</sup>C<sub>0</sub> $\times$ <sup>7</sup>C<sub>4</sub>
<br><br>= 70 + 210 + 175 + 35
<br><br>= 490
About this question
Subject: Mathematics · Chapter: Permutations and Combinations · Topic: Fundamental Counting Principle
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