Permutations & Combinations
Combinations
Grade 11
Question:
<p>A shopkeeper places before you 41 different toys out which 20 toys are to be purchased. Suppose \(m\) = number of ways in which 20 toys can be purchased without any restriction and \(n\) = number of ways in which a particular toy is to be always included in each selection of 20 toys, then \((m - n)\) can be expressed as:</p>
<p>(a) \(\dfrac{2^{10}}{20!}(1 \cdot 3 \cdot 5 \cdots 39)\)</p>
<p>(b) \(\dfrac{2^{20}(1 \cdot 3 \cdot 5 \cdots 19)}{10!}\)</p>
<p>(c) \(\displaystyle\prod_{r=0}^{19}\left(\dfrac{4r+2}{20-r}\right)\)</p>
<p>(d) \(\left(\dfrac{21}{1}\right)\left(\dfrac{22}{2}\right)\left(\dfrac{23}{3}\right)\cdots\left(\dfrac{40}{20}\right)\)</p>
Step-by-Step Solution
Key Concept: The difference (m - n) represents selections where a particular toy is NOT included; this equals choosing 20 toys from the remaining 40 toys, which is C(40,20). Recognize that m - n = C(40,20) directly without computing m and n separately.
<p><strong>Step 1: Define m and n</strong></p><p>m = number of ways to select 20 toys from 41 toys without restriction = C(41,20)</p><p>n = number of ways to select 20 toys such that a particular toy is always included = C(40,19)</p><p>(Since 1 toy is fixed, choose remaining 19 from 40 toys)</p><p><strong>Step 2: Interpret (m - n) combinatorially</strong></p><p>m - n represents selections of 20 toys where the particular toy is <strong>NOT</strong> included.</p><p>This directly equals: C(40,20) = number of ways to select 20 toys from the remaining 40 toys</p><p><strong>Step 3: Verify using algebraic identity</strong></p><p>m - n = C(41,20) - C(40,19)</p><p>Using C(n,r) - C(n-1,r-1) = C(n-1,r):</p><p>C(41,20) - C(40,19) = C(40,20) ✓</p><p><strong>Step 4: Express the answer</strong></p><p>(m - n) = C(40,20) = 40!/(20!·20!)</p><p>∴ Answer: ACD</p>
Correct Answer: ACD