<p>A double-decker bus carry \((u + \ell)\) passengers, \(u\) in the upper deck and \(\ell\) in the lower deck. Find the number of ways in which the \((u + \ell)\) passengers can be distributed in the two decks, if \(r\ (\leq \ell)\) particular passengers refuse to go in the upper deck and \(s\ (\leq u)\) refuse to sit in the lower deck.</p>
Step-by-Step Solution
Key Concept: The constrained passengers must occupy fixed positions (r in lower, s in upper), so only the remaining (u+ℓ-r-s) unrestricted passengers can be freely distributed between the two decks. This becomes a problem of choosing which of the remaining passengers go to the upper deck.
<p><strong>Step 1: Identify fixed placements</strong></p><p>• r particular passengers <strong>refuse upper deck</strong> → they MUST go to lower deck<br>• s particular passengers <strong>refuse lower deck</strong> → they MUST go to upper deck<br>• These are already placed with no choices</p><p><strong>Step 2: Count remaining unrestricted passengers</strong></p><p>Total passengers = u + ℓ<br>Constrained passengers = r + s<br>Unrestricted passengers = (u + ℓ - r - s)</p><p><strong>Step 3: Determine capacity constraints for unrestricted passengers</strong></p><p>Upper deck: Already has s passengers (forced), can accommodate (u - s) more<br>Lower deck: Already has r passengers (forced), can accommodate (ℓ - r) more</p><p><strong>Step 4: Count valid distributions</strong></p><p>From the (u + ℓ - r - s) unrestricted passengers, we must choose how many go to the upper deck.<br>If k passengers go to upper: k ≤ (u - s) and (u + ℓ - r - s - k) ≤ (ℓ - r)<br>This simplifies to: 0 ≤ k ≤ (u - s)<br><br>Number of ways = $$\sum_{k=0}^{u-s} \binom{u+\ell-r-s}{k}$$<br>By the hockey-stick identity: $$\sum_{k=0}^{u-s} \binom{u+\ell-r-s}{k} = \binom{u+\ell-r-s+1}{u-s}$$<br><br>This equals: $$\dfrac{(u+\ell-r-s)!}{(\ell-r)!(u-s)!}$$</p><p>∴ Answer: $$\dfrac{(u+\ell-r-s)!}{(\ell-r)!(u-s)!}$$</p>
Correct Answer: \(\dfrac{(u+\ell-r-s)!}{(\ell-r)!(u-s)!}\)