Question

(c) use the usual method to construct row 11 in pascals triangle. verify that the same pattern holds. select the correct choice below and, if necessary, fill in the answer boxes to complete your choice.

a. row 11 has 11 values.
row 11

b. row 11 has 10 values.
row 11

c. row 11 has 12 values.
row 11

d. row 11 cannot be constructed.

Explanation:

Step1: Recall Pascal's triangle property

The \(n\) -th row of Pascal's triangle has \(n + 1\) elements. For \(n=11\), the number of elements in row 11 is \(11 + 1=12\).

Step2: Construct row 11 using binomial coefficients

The elements of the \(n\) -th row of Pascal's triangle are given by the binomial coefficients \(\binom{n}{k}\) for \(k = 0,1,\cdots,n\). For \(n = 11\), we calculate:
\(\binom{11}{0}=1\), \(\binom{11}{1}=\frac{11!}{1!(11 - 1)!}=\frac{11!}{1!10!}=11\), \(\binom{11}{2}=\frac{11!}{2!(11 - 2)!}=\frac{11\times10}{2\times 1}=55\), \(\binom{11}{3}=\frac{11!}{3!(11 - 3)!}=\frac{11\times10\times9}{3\times2\times1}=165\), \(\binom{11}{4}=\frac{11!}{4!(11 - 4)!}=\frac{11\times10\times9\times8}{4\times3\times2\times1}=330\), \(\binom{11}{5}=\frac{11!}{5!(11 - 5)!}=\frac{11\times10\times9\times8\times7}{5\times4\times3\times2\times1}=462\), \(\binom{11}{6}=\binom{11}{5}=462\) (due to the symmetry of binomial coefficients \(\binom{n}{k}=\binom{n}{n - k}\)), \(\binom{11}{7}=\binom{11}{4}=330\), \(\binom{11}{8}=\binom{11}{3}=165\), \(\binom{11}{9}=\binom{11}{2}=55\), \(\binom{11}{10}=\binom{11}{1}=11\), \(\binom{11}{11}=\binom{11}{0}=1\).

Answer:

C. Row 11 has 12 values.
Row 11 1 11 55 165 330 462 462 330 165 55 11 1