Permutations & Combinations
Grade None

Question:

<p>The total number of functions, f:{1, 2, 3, 4}&nbsp;<span class="math-tex">\(\rightarrow\)</span> {1, 2, 3, 4, 5, 6} such that f(1)&nbsp;+ f(2) = f(3), is equal to:</p>
<p style="display:inline">90</p>
<p style="display:inline">108</p>
<p style="display:inline">126</p>
<p style="display:inline">60</p>

Step-by-Step Solution

Key Concept: Solve the problem by enumerating possible values for f(3) to determine valid pairs of (f(1), f(2)) and then account for the six independent choices for the domain element f(4).
<p>Given A = {1, 2, 3, 4}<br /> B = {1, 2, 3, 4, 5, 6}<br /> Here&nbsp;f(3) can be&nbsp;2, 3, 4, 5, 6<br /> f(3) = 2, (f(1), f(2)) <span class="math-tex">$\rightarrow$</span>(1, 1) <span class="math-tex">$\rightarrow$</span> Total&nbsp;6 cases<br /> f(3) = 3, (f(1), f(2)) <span class="math-tex">$\rightarrow$</span>&nbsp;(1, 2), (2, 1)<br /> <span class="math-tex">$\rightarrow$</span>&nbsp;2 <span class="math-tex">$\times$</span> 6 = Total 12 cases<br /> f(3) = 4,(f(1), f(2)) <span class="math-tex">$\rightarrow$</span>(1, 3), (3, 1), (2, 2)<br /> <span class="math-tex">$\rightarrow$</span>3 <span class="math-tex">$\times$</span> 6 =&nbsp;Total&nbsp;18 cases<br /> f(3) = 5, (f(1), f(2)) <span class="math-tex">$\rightarrow$</span>&nbsp;(1, 4), (4, 1), (2, 3), (3, 2)<br /> <span class="math-tex">$\rightarrow$</span> 4 <span class="math-tex">$\times$</span> 6 = Total&nbsp;24 cases<br /> f(3) = 6,(f(1), f(2)) <span class="math-tex">$\rightarrow$</span>&nbsp;(1, 5), (5, 1), (2, 4), (4, 2), (3, 3)<br /> <span class="math-tex">$\rightarrow$</span> 5 <span class="math-tex">$\times$</span> 6 =&nbsp;Total&nbsp;30 cases<br /> Total number of cases = 6 + 12 + 18 + 24 + 30 = 90</p>
Correct Answer: A

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