Question

Question: a. Determine the number of k-faces of the 5-dimensional hypercube \({C^{\bf{5}}}\) for \(k = {\bf{0}},{\bf{1}},.....,{\bf{4}}\). Verfy that your answer satisfies Euler’s formula.

b. Make a chart of the values of \({f_k}\left( {{C^n}} \right)\) for \(n = {\bf{1}},.....,{\bf{5}}\) and \(k = {\bf{0}},{\bf{1}},.....,{\bf{4}}\). Can you see a pattern? Guess a general formula for \({f_k}\left( {{C^n}} \right)\).

Short answer

a. 32, 80, 80, 40, 10

b.

	\({f_0}\)	\({f_1}\)	\({f_3}\)	\({f_4}\)	\({f_5}\)
\({S^1}\)	2
\({S^2}\)	4	4
\({S^3}\)	8	12	6
\({S^4}\)	16	32	24	8
\({S^5}\)	32	80	80	40	10

There exist a pattern for \({f_k}\left( {{S^n}} \right)\) and it is given by the formula, \({f_k}\left( {{C^n}} \right) = {2^{k + 1}}\left( {\begin{array}{*{20}{c}}n\\k\end{array}} \right)\). Here, \(\left( {\begin{array}{*{20}{c}}a\\b\end{array}} \right) = \frac{{a!}}{{b!\left( {a - b} \right)!}}\) is the binomial coefficient.

Step-by-step solution

Find solution for part (a)

The number offaces for \(k = 0\):

\({f_0}\left( {{C^5}} \right) = 32\)

The number of faces for \(k = 1\):

\({f_1}\left( {{C^5}} \right) = 80\)

The number of faces for \(k = 2\):

\({f_2}\left( {{C^5}} \right) = 80\)

The number of faces for \(k = 3\):

\({f_3}\left( {{C^5}} \right) = 40\)

The number of faces for \(k = 4\):

\({f_4}\left( {{C^5}} \right) = 10\)

Verify the Euler’s formula

Euler’s formulacan be verified as follows:

\(\begin{array}{c}{f_0}\left( {{C^5}} \right) - {f_1}\left( {{C^5}} \right) + {f_2}\left( {{C^5}} \right) - {f_3}\left( {{C^5}} \right) + {f_4}\left( {{C^5}} \right) = 32 - 80 + 80 - 40 + 10\\ = 2\end{array}\)

So, Euler’s formula is verified.

Find the solution for part (b)

The chart below represents the values of \({f_k}\left( {{C^n}} \right)\).

	\({f_0}\)	\({f_1}\)	\({f_3}\)	\({f_4}\)	\({f_5}\)
\({S^1}\)	2
\({S^2}\)	4	4
\({S^3}\)	8	12	6
\({S^4}\)	16	32	24	8
\({S^5}\)	32	80	80	40	10

There exist a pattern for the values in the chart. The pattern for the above chart is given by the formula \({f_k}\left( {{C^n}} \right) = {2^{k + 1}}\left( {\begin{array}{*{20}{c}}n\\k\end{array}} \right)\). Here \(\left( {\begin{array}{*{20}{c}}a\\b\end{array}} \right) = \frac{{a!}}{{b!\left( {a - b} \right)!}}\) is the binomial coefficient.