# Would you show how many different orders in this case is possible?

You want to arrange your disks so that the cd's of the same artist come in a row and you have 20 cd's from five different artists four from each. How many different orders in this case is possible?

My solution

Firstly the cd's of each artist can be arranged $4 \cdot 4!=96$ different ways. In addition each artist can be arranged $5!=120$ different ways. Thus by the principle of product in combinatorics the total number of different orders is $96 \cdot 120 = 11520$. I would like to ask if this is right result and if you wish to show your own result?

-

## 1 Answer

Since $4$ cd's can be arranged in $4!$ different ways and $5$ artists can be arranged in $5!$ different ways I think that solution is given by :

$5! \cdot (4! \cdot 4! \cdot 4! \cdot 4! \cdot 4! ) = 5! \cdot (4!)^5$ different ways .

-

In my calculations I got $5! \cdot 5 \cdot 4! = 12600$. Are you sure that it is 14400? No I got the same result. It is $120 \cdot 120= 14400$ – alvoutila Feb 13 '12 at 11:09
@alvoutila,$120 \cdot 5 \cdot 24$ – pedja Feb 13 '12 at 11:12
@AndréNicolas,Of course...thanks.. – pedja Feb 13 '12 at 17:12
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 11, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9440415501594543, "perplexity_flag": "middle"}