The number of ordered pairs (r, k) for which
6.35Cr
= (k2 - 3). 36Cr + 1, where k is an integer, is :
Solution
6.<sup>35</sup>C<sub>r</sub>
= (k<sup>2</sup> - 3). <sup>36</sup>C<sub>r + 1</sub>
<br><br>$\Rightarrow$ 6.<sup>35</sup>C<sub>r</sub>
= (k<sup>2</sup> - 3).${{36} \over {r + 1}}$<sup>35</sup>C<sub>r</sub>
<br><br>$\Rightarrow$ k<sup>2</sup> - 3 = ${{r + 1} \over 6}$
<br><br>Possible values of r for integral values of k, are
<br><br>r = 5, 35
<br><br>number of ordered pairs are 4
<br><br>(5, 2), (5, –2), (35, 3), (35, 3)
About this question
Subject: Mathematics · Chapter: Permutations and Combinations · Topic: Fundamental Counting Principle
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