Consider a rectangle ABCD having 5, 7, 6, 9 points in the interior of the line segments AB, CD, BC, DA respectively. Let $\alpha$ be the number of triangles having these points from different sides as vertices and $\beta$ be the number of quadrilaterals having these points from different sides as vertices. Then ($\beta$ $-$ $\alpha$) is equal to :
Solution
$$\alpha = {}^6{C_1}{}^7{C_1}{}^9{C_1} + {}^5{C_1}{}^7{C_1}{}^9{C_1} + {}^5{C_1}{}^6{C_1}{}^9{C_1} + {}^5{C_1}{}^6{C_1}{}^7{C_1} $$
<br><br>$= 378 + 315 + 270 + 210 = 1173$<br><br>$\beta = {}^5{C_1}{}^6{C_1}{}^7{C_1}{}^9{C_1} = 1890$<br><br>$\therefore$ $\beta - \alpha = 1890 - 1173 = 717$
About this question
Subject: Mathematics · Chapter: Permutations and Combinations · Topic: Fundamental Counting Principle
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