An-8-8-checkerboard-is-to-be-coloured-with-6-colors-in-a-way-that-no-2-adjacent-cells-sharing-side-have-the-same-color-In-how-many-ways-can-it-be-done-

Question Number 877 by sagarwal last updated on 07/Apr/15

An 8 \times 8 checkerboard is to be coloured

with 6 colors in a way that no 2 adjacent

cells (sharing side) have the same color . In how

many ways can it be done .

Commented by 123456 last updated on 12/Apr/15

[\begin{matrix} 6 & 5 & 5 & 5 & 5 & 5 & 5 & 5 \\ 5 & 4 & 4 & 4 & 4 & 4 & 4 & 4 \\ 5 & 4 & 4 & 4 & 4 & 4 & 4 & 4 \\ 5 & 4 & 4 & 4 & 4 & 4 & 4 & 4 \\ 5 & 4 & 4 & 4 & 4 & 4 & 4 & 4 \\ 5 & 4 & 4 & 4 & 4 & 4 & 4 & 4 \\ 5 & 4 & 4 & 4 & 4 & 4 & 4 & 4 \\ 5 & 4 & 4 & 4 & 4 & 4 & 4 & 4 \end{matrix}]

k = 6 \times 5^{7 + 7} \times 4^{7 \times 7}

k = 6 \times 5^{14} \times 4^{49}

k = (2 \times 3) \times 5^{14} \times {(2^{2})}^{49}

k = 2 \times 3 \times 5^{14} \times 2^{2 \times 49}

k = 2 \times 3 \times 5^{14} \times 2^{98}

k = 2^{98 + 1} \times 3 \times 5^{14}

k = 2^{99} \times 3 \times 5^{14}

\log k = \log (2^{99} \times 3 \times 5^{14})

\log k = 99 \log 2 + \log 3 + 14 \log 5

\log k = \frac{99 \ln 2 + \ln 3 + 14 \ln 5}{\ln 10}

\log k \approx 40, 06

40 ⪅ \log k < 41

10^{40} ⪅ k < 10^{41}

Commented by prakash jain last updated on 13/Apr/15

If we have only 2 color the grid approach

fills with 0 s . I think what is missing are

special cases when all adjacent cells are

filled with same color as in 2 - color cases .