Question

$6$ Letters are to be posted in three boxes. The numbers of ways of posting the letters when no letterbox remains empty is
A. $270\;$
B. $540\;$
C. $537\;$
D. None of these

Answer 1

The number of letters that are to be posted in three boxes is $6$.So, write the possible number of ways the letters can be put into boxes without repetition order. Now find the permutations and combinations for those orders and then finally add them to get the number of ways of posting letters such that no letterbox remains empty.

Formula used: The number of permutations of n objects taken r at a time is, $_n{p_r} = \dfrac{{n!}}{{(n - r)!}}$
Permutations with repetitions: out of n objects in a set, p are exactly alike of one kind, q are exactly alike of one kind, r are exactly alike of one kind then, $= \dfrac{{n!}}{{p!q!r!}}$

Complete step-by-step solution:
Given that there are three boxes and six letters.
The possible ways to put them without leaving any box empty are:
$(1,1,4);(1,2,3);(2,2,2)$
There are three ways to do so without repetition.
Now let’s take the first condition where there are $1,1,4\;$ letters respectively.
We use the formula for permutations. Which is, Permutations with repetitions: out of n objects in a set, p are exactly alike of one kind, q are exactly alike of one kind, r are exactly alike of one kind then, $= \dfrac{{n!}}{{p!q!r!}}$
The permutation can we write as,
$\Rightarrow \dfrac{{6!}}{{4!1!1!2!}}$ Here we write $2!\;$ since $1$ is repeated two times.
On evaluating, we expand the factorial.
$\Rightarrow \dfrac{{6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{4 \times 3 \times 2 \times 2}}$
On simplifying we get,
$\Rightarrow \dfrac{{3 \times 5 \times 1}}{1} = 15$
Now the second condition where there are $1,2,3\;$ letters respectively.
The permutation can we write as,
$\Rightarrow \dfrac{{6!}}{{1!2!3!}}$
On evaluating, we expand the factorial.
$\Rightarrow \dfrac{{6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{1 \times 2 \times 3 \times 2}}$
On simplifying we get,
$\Rightarrow \dfrac{{6 \times 5 \times 2}}{1} = 60$
Now the third condition where there are $2,2,2\;$ letters respectively.
The permutation can we write as,
$\Rightarrow \dfrac{{6!}}{{2!2!2!3!}}$ Here we write $3!\;$ since $2$ is repeated three times.
On evaluating, we expand the factorial.
$\Rightarrow \dfrac{{6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 2 \times 2 \times 3 \times 2}}$
On simplifying we get,
$\Rightarrow \dfrac{{3 \times 5 \times 1}}{1} = 15$
Since we are taking the order of boxes without repetition the number of ways of arranging the boxes are $3!\;$
Now the total ways can we write as $(15 + 60 + 15) \times 3!$
On solving we will get,
$\Rightarrow 90 \times 3!$
$\Rightarrow 90 \times 6$
Which will be equal to the value,
$\Rightarrow 540$
$\therefore$ There are $540\;$ ways to arrange $6$ letters in $3$ boxes.

Option B is the correct answer.

Note: Whenever there are two or more quantities (who are the same of one kind) that are equal in number, then multiply the factorial of the number of kinds in the denominator also along with the other permutations.