Permutations & Combinations
Counting Solutions
Grade 11
Question:
<p>Let <span class='math'>x</span> be the elements of the set <span class='math'>A = \{1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120\}</span> and <span class='math'>x_1, x_2, x_3</span> be positive integers and <span class='math'>d</span> be the number of integral solutions of <span class='math'>x_1 x_2 x_3 = x</span>, then <span class='math'>d</span> is</p>
Step-by-Step Solution
Key Concept: The number of ordered factorizations of n into three factors depends on the prime factorization of n.
<p>For each element <span class='math'>x \in A</span>, we need to count ordered triples <span class='math'>(x_1, x_2, x_3)</span> such that <span class='math'>x_1 x_2 x_3 = x</span>. This is equivalent to finding the number of divisors of each element when expressed as ordered products of three factors.</p><p>Using the formula for ordered factorizations: if <span class='math'>x = p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k}</span>, the number of ordered triples is <span class='math'>\binom{a_1+2}{2} \binom{a_2+2}{2} \cdots \binom{a_k+2}{2}</span>.</p><p>Computing for all elements in <span class='math'>A</span> and summing gives <span class='math'>d = 150</span>.</p>
Correct Answer: 150