Question

A professor packs her collection of 40 issues of a mathematics journal in four boxes with 10 issues per box. How many ways can she distribute the journals if
a) each box is numbered, so that they are distinguishable?
b) the boxes are identical, so that they cannot be distinguished?

Answer 1

Answer:

a) $\frac{40!}{(10!)^4}$

b) $\frac{40!}{(10!)^4 4!}$

Step-by-step explanation:

Given : A professor packs her collection of 40 issues of a mathematics journal in four boxes with 10 issues per box.

To find : How many ways can she distribute the journals if

a) each box is numbered, so that they are distinguishable?

b) the boxes are identical, so that they cannot be distinguished?

Solution :

A professor packs her collection of 40 issues of a mathematics journal in four boxes with 10 issues per box.

a) each box is numbered, so that they are distinguishable

The first can filled in $^{40}C_{10}$ ways.

After the 1st box filled we are left with 30 journals.

The second can filled in $^{30}C_{10}$ ways.

After the 2nd box filled we are left with 20 journals.

The third can filled in $^{20}C_{10}$ ways.

After the 3rd box filled we are left with 10 journals.

The fourth can filled in $^{10}C_{10}$ ways.

The journals can be distributed in

$^{40}C_{10}\times^{30}C_{10}\times^{20}C_{10}\times ^{10}C_{10}$

$=\frac{40!}{(40-10)!10!}\times\frac{30!}{(30-10)!10!}\times\frac{20!}{(20-10)!10!}\times \frac{10!}{(10-10)!10!}$

$=\frac{40!}{30!10!}\times\frac{30!}{20)!10!}\times\frac{20!}{10!10!}\times \frac{10!}{0!10!}$

$=\frac{40!}{(10!)^4}$

b) the boxes are identical, so that they cannot be distinguished?

Suppose that four boxes are identical.

Since the four boxes are identical, they can be arranges in 4! ways.

As 40 issues of mathematical journal can be distributed in to four boxes with 10 issues per box in $\frac{40!}{(10!)^4}$ ways.

Finally journal can be distributed into four identical boxes with 10 in each box is $\frac{40!}{(10!)^4 4!}$ ways.