Answer to Question #134531 in Combinatorics | Number Theory for boomCarti

Question #134531
How many words can we build using exactly 4 A's, 4 B's and 4 C's if the first 4 letters cannot be A's, the second 4 letters cannot be B's and the third 4 letters cannot be C's?
1
Expert's answer
2020-09-23T18:19:36-0400

Assume that exactly "N" A’s stand on the last 4 positions (so "\ud835\udc41\u2208\\{0,1,2,3,4\\}" ). It follows that there are "4\u2212\ud835\udc41" B’s on the last 4 positions and "4\u2212\ud835\udc41" A’s on the middle 4 positions. Also it follows that there are "\ud835\udc41" C’s on the middle 4 positions and "4-\ud835\udc41" C’s on the first 4 positions.

So there are "\\begin{pmatrix} 4 \\\\N \\end{pmatrix}" ways to place "N" A’s on the last 4 positions. Then there are "4-N" ways to place the rest A’s on the middle 4 positions. Finally, there are "\\begin{pmatrix} 4 \\\\N \\end{pmatrix}" ways to place "N" B’s on the first 4 positions. So if we place "N" A’s on the last 4 positions, "4-N" A’s on the middle 4 positions and "N" B’s on the first 4 positions, then the rest positions can be filled in the only way –C’s on the first and on the middle 4 positions, B’s on the last 4 positions.


Hence the number of words with "N" A’s on the last 4 positions is equal to

"\\begin{pmatrix} 4 \\\\N \\end{pmatrix} \\cdot \\begin{pmatrix} 4 \\\\4-N \\end{pmatrix} \\cdot \\begin{pmatrix} 4 \\\\N \\end{pmatrix} = \\begin{pmatrix} 4 \\\\N \\end{pmatrix}^3"

and the total number is

"\\displaystyle \\sum_{N=0}^4 \\begin{pmatrix} 4 \\\\N \\end{pmatrix}^3 = 1^3 + 4^3 + 6^3+4^3+1^3=1+64+216+64+1=346"

Answer: 346


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