<p><strong>Example 94:</strong> Find the number of different selections of 5 letters which can be made from 5A's, 4B's, 3C's, 2D's and 1E.</p>
Step-by-Step Solution
Key Concept: Use generating functions where each factor represents available choices of each item. The coefficient of the desired power gives the number of selections.
<p><strong>Solution:</strong> All selections of 5 letters are given by the 5th degree terms in the generating function:</p><p>$$(1 + A + A^2 + A^3 + A^4 + A^5)(1 + B + B^2 + B^3 + B^4)(1 + C + C^2 + C^3)(1 + D + D^2)(1 + E)$$</p><p>The number of 5-letter selections equals the coefficient of $\alpha^5$ in:</p><p>$$(1 + \alpha + \alpha^2 + \alpha^3 + \alpha^4 + \alpha^5)(1 + \alpha + \alpha^2 + \alpha^3 + \alpha^4)(1 + \alpha + \alpha^2 + \alpha^3)(1 + \alpha + \alpha^2)(1 + \alpha)$$</p><p><strong>Using synthetic multiplication (adding preceding coefficients):</strong></p><p>Step 1: Multiply by $(1 + \alpha)$ — add 1 preceding coefficient<br/>Coefficients: 1, 2, 3, 4, 5, 5, ...</p><p>Step 2: Multiply by $(1 + \alpha + \alpha^2)$ — add 2 preceding coefficients<br/>Coefficients: 1, 3, 6, 10, 14, 17, ...</p><p>Step 3: Continue with $(1 + \alpha + \alpha^2 + \alpha^3)$ and $(1 + \alpha + \alpha^2 + \alpha^3 + \alpha^4)$ — add 3 and 4 preceding coefficients respectively<br/>Final coefficients: 1, 5, 12, 19, 22, 19, 12, ...</p><p>Step 4: Multiply by $(1 + \alpha + \alpha^2 + \alpha^3 + \alpha^4 + \alpha^5)$ — add 5 preceding coefficients<br/>Coefficient of $\alpha^5$: 1, 4, 7, 8, 7, 4, 1, ...</p><p>∴ <strong>Coefficient of $\alpha^5$ = 53</strong></p>
Correct Answer: 53