Question

There are$3$ apartments A, B and C for rent in a building. Each apartment will accept either $3$ or$4$ occupants. The number of ways of renting the apartments to $10$ students
A) 12600
B) 10800
C) 13500
D) 15000

Answer 1

Hint: To solve this question we need to know the concept of combination. ${}^{a}{{C}_{b}}$ in the terms of combination could be written as $\dfrac{a!}{b!\left( a-b \right)!}$ where $a!$ means product of all numbers from $1$ to $a$, including $a$, mathematically $a!=a\times \left( a-1 \right)\times \left( a-2 \right)\times ............\times 2\times 1$.

Complete step by step solution:
The question asks you to find the total number of ways in which$10$ students can accommodate themselves in three apartments, namely A, B and C in which either $3$or $4$occupants can be accommodated in each apartment.
To solve this problem we need to analyse the condition that each apartment should have either of $3$ or $4$ members. The students are that can be accommodated can be in ways:
First way:
$4$ in Apartment $A$+$3$ in Apartment $B$+$3$ in Apartment $C$
Second way:
$3$ in Apartment $A$+$3$ in Apartment $B$+$4$ in Apartment $C$
Third way:
$3$ in Apartment $A$+$4$ in Apartment $B$+$3$ in Apartment $C$
To solve the problem, when $10$ students are accommodated as per the first way which is $4$ in Apartment $A$+$3$ in Apartment $B$+$3$ in Apartment $C$ we get:
$= {}^{10}{{C}_{4}}\times {}^{6}{{C}_{3}}\times {}^{3}{{C}_{3}}$
\[= \dfrac{10!}{4!6!}\times \dfrac{6!}{3!3!}\times \dfrac{3!}{3!0!}\]
Some of the formulas used to find the value of the above factorial is $n!=n\times \left( n-1 \right)\times \left( n-2 \right)......\times 3\times 2\times 1,0!=1$ and $1!=1$
On solving this factorial using the above formula we get:
$= \dfrac{10\times 9\times 8\times 7\times 6!}{4!6!}\times \dfrac{6\times 5\times 4\times 3!}{3!3!}\times 1$
$= \dfrac{10\times 9\times 8\times 7}{4\times 3\times 2\times 1}\times \dfrac{6\times 5\times 4}{3\times 2\times 1}\times 1$
On cancelling the terms, which are common in numerator and denominator, we get:
$= 10\times 3\times 4\times 7\times 5$
$= $$4200$
Similarly for the second way in which accommodation is in this way
$3$in Apartment $A$+$3$in Apartment $B$+$4$in Apartment $C$, so number of ways are
$= {}^{10}{{C}_{3}}\times {}^{7}{{C}_{3}}\times {}^{4}{{C}_{4}}$
\[= \dfrac{10!}{3!7!}\times \dfrac{7!}{3!4!}\times \dfrac{4!}{4!0!}\]
$= \dfrac{10\times 9\times 8\times 7!}{7!3!}\times \dfrac{7\times 6\times 5\times 4!}{3!4!}\times 1$
$= \dfrac{10\times 9\times 8}{3\times 2\times 1}\times \dfrac{7\times 6\times 5}{3\times 2\times 1}\times 1$
$= \dfrac{10\times 3\times 4}{1}\times \dfrac{7\times 5}{1}\times 1$
$= 4200$
Now for, the third way which says $3$ in Apartment $A$+$4$ in Apartment $B$+$3$ in Apartment $C$
$= {}^{10}{{C}_{3}}\times {}^{7}{{C}_{4}}\times {}^{3}{{C}_{3}}$
\[= \dfrac{10}{3!7!}\times \dfrac{7!}{3!4!}\times \dfrac{3!}{3!0!}\]
$= \dfrac{10\times 9\times 8\times 7!}{3!7!}\times \dfrac{7\times 6\times 5\times 4!}{3!4!}\times 1$
$= \dfrac{10\times 9\times 8}{3\times 2\times 1}\times \dfrac{7\times 6\times 5}{3\times 2\times 1}\times 1$
$= \dfrac{10\times 3\times 4}{1}\times \dfrac{7\times 5}{1}\times 1$
$= 4200$
On adding all ways three that together we get:
$= 4200+4200+4200$
$= 12600$
$\therefore $ The total number of ways in which $10$ students can be accommodated in $3$ apartments are $A)12600$ .

So, the correct answer is “Option A”.

Note: We solved the question by taking the three cases individually, which made the solution a bit lengthy. Instead we would have calculated the ways for single arrangement and then would have multiplied it by $3$ . This is done because all the three arrangements are equivalent to each other. So in a shorter manner $4200$ would have been multiplied by $3$ fetching the number of ways to be $12600$ .