Question

How many ways are there to distribute six objects to five boxes if

a) Both the objects and boxes are labeled?

b) The objects are labeled, but the boxes are unlabeled?

c) The objects are unlabeled, but the boxes are labeled?

d) Both the objects and the boxes are unlabeled?

Short answer

(a)The number of ways to six objects distribute to five boxes is\(15625\).

(b) The number of ways to distribute six objects to five boxes if objects are labeled but the boxes is unlabeled.

(c)The required number of ways distribute six objects to five boxes if objects are unlabeled and boxes are labeled is \(210\) ways.

(d) The number of ways to distribute six objects to five boxes if both the objects and boxes are unlabeled is\(10\).

Step-by-step solution

Find labeled boxes.

Six objects and five boxes are given.

(a) Here each object distribute to five boxes in five ways, hence by product rule.

The number of ways to distribute six objects to five boxes is\({5^6} = 15625\).

Find unlabeled boxes.

(b) Since objects are labeled but the boxes are unlabeled, we can use the sterling number formula.

\({\rm{i}}{\rm{.e}}{\rm{.,}}\sum\limits_{{\rm{j = 1}}}^{\rm{k}} {\frac{{\rm{1}}}{{{\rm{j!}}}}} \sum\limits_{{\rm{i = 0}}}^{{\rm{j - 1}}} {{{{\rm{( - 1)}}}^{\rm{i}}}} {\rm{c(ji)(j - i}}{{\rm{)}}^{\rm{n}}}\)

Where n=6 and k=5, we get. \(\begin{array}{c}\frac{{\rm{1}}}{{{\rm{1!}}}}{{\rm{1}}^{\rm{6}}}{\rm{ + }}\frac{{\rm{1}}}{{{\rm{2!}}}}{\rm{(1}}{\rm{.}}{{\rm{2}}^{\rm{6}}}{\rm{ - 2}}{\rm{.}}{{\rm{1}}^{\rm{6}}}{\rm{) + }}\frac{{\rm{1}}}{{{\rm{3!}}}}{\rm{(1}}{\rm{.}}{{\rm{3}}^{\rm{6}}}{\rm{ - 3}}{\rm{.}}{{\rm{2}}^{\rm{6}}}{\rm{ + 3}}{\rm{.}}{{\rm{1}}^{\rm{6}}}{\rm{) + }}\frac{{\rm{1}}}{{{\rm{4!}}}}{\rm{(1}}{\rm{.}}{{\rm{4}}^{\rm{6}}}{\rm{ - 4}}{\rm{.}}{{\rm{3}}^{\rm{6}}}{\rm{ + 6}}{\rm{.}}{{\rm{2}}^{\rm{6}}}{\rm{ - 4}}{\rm{.}}{{\rm{1}}^{\rm{6}}}{\rm{) + }}\frac{{\rm{1}}}{{{\rm{5!}}}}{\rm{(1}}{\rm{.}}{{\rm{5}}^{\rm{6}}}{\rm{ - 5}}{\rm{.}}{{\rm{4}}^{\rm{6}}}{\rm{ + 10}}{\rm{.}}{{\rm{3}}^{\rm{6}}}{\rm{ - 10}}{\rm{.}}{{\rm{2}}^{\rm{6}}}{\rm{ + 5}}{\rm{.}}{{\rm{1}}^{\rm{6}}}{\rm{)}}\\{\rm{ = 1 + 31 + 90 + 65 + 15 = 202}}\end{array}\)

Therefore, the number of ways to distribute six objects to five boxes if objects are labeled but the boxes are unlabeled is\({\rm{201}}\).

Find both unlabeled and labeled boxes.

(c) This case is similar to choosing an n-combination from the set of \({\rm{k}}\) boxes with repetition allowed is\({\rm{c(n + k - 1,k - 1)}}\).

Where \({\rm{n = 6}}\) and \({\rm{k = 5}}\)

Therefore, the required number of ways distribute six objects to five boxes if objects are unlabeled and boxes are labeled is\(\begin{array}{c}{\rm{c(6 + 5 - 1,5 - 1) = }}\\{\rm{c(10,4) = }}\frac{{{\rm{10 \times 9 \times 8 \times 7}}}}{{{\rm{1 \times 2 \times 3 \times 4}}}}{\rm{ = 210ways}}{\rm{.}}\end{array}\)

Find unlabeled boxes and object.

Since both the boxes and objects are indistinguishable, which is nothing but how many different ways there are to write 6 as sum of five nonnegative integers without order is

\(\begin{array}{c}{\rm{6 = 6 + 0 + 0 + 0 + 0;6 = 5 + 1 + 0 + 0 + 0;}}\\{\rm{6 = 4 + 2 + 0 + 0 + 0;6 = 4 + 1 + 1 + 0 + 0;}}\\{\rm{6 = 3 + 3 + 0 + 0 + 0;6 = 3 + 2 + 1 + 0 + 0;}}\\{\rm{6 = 3 + 1 + 1 + 1 + 0;6 = 2 + 2 + 2 + 0 + 0;}}\\{\rm{6 = 2 + 2 + 1 + 1 + 0;6 = 2 + 1 + 1 + 1 + 1}}\end{array}\)

Thus there are \(10\) ways to write it.

Hence the number of ways to distribute six objects to five boxes if both the objects and boxes are unlabeled is\(10\).