Question

If a, b and c are the greatest values of $^{19}C_{p}, \, ^{20}C_{q}$ and $^{21}C_{r}$ respectively, then :

• A. $\frac{a}{11} = \frac{b}{22} = \frac{c}{42}$
• B. $\frac{a}{10} = \frac{b}{11} = \frac{c}{42}$
• C. $\frac{a}{11} = \frac{b}{22} = \frac{c}{21}$
• D. $\frac{a}{10} = \frac{b}{11} = \frac{c}{21}$

Answer 1

The Correct Option is A

 To determine the greatest values of \(_nC_k\), we use the concept that a binomial coefficient \(_nC_k\) is maximized when \(k\) is approximately half of \(n\). This is because the central binomial coefficient is the largest.

1. Finding the value of \(a\):

The greatest value of \(_{19}C_p\) occurs at \(p\) close to 9 or 10. Since \(_{19}C_9 = _{19}C_{10}\), we take \(_{19}C_9\) as the greatest binomial coefficient for simplicity.

Therefore, \(a = _{19}C_9\).

1. Finding the value of \(b\):

The greatest value of \(_{20}C_q\) occurs at \(q = 10\), which is exactly half of 20.

Therefore, \(b = _{20}C_{10}\).

1. Finding the value of \(c\):

The greatest value of \(_{21}C_r\) occurs at \(r = 10\) or \(r = 11\). Both \(_{21}C_{10}\) and \(_{21}C_{11}\) are equal and maximum in value.

Therefore, \(c = _{21}C_{10}\) (or equivalently \(_{21}C_{11}\)).

To solve for the given problem, we need to compare \(a\), \(b\), and \(c\) according to the options.

Substituting these values in the expressions:

1. Calculate \(_{19}C_9\):

\(_{19}C_9 = \frac{19 \times 18 \times 17 \times 16 \times 15 \times 14 \times 13 \times 12 \times 11}{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}\) simplifies to calculate as:

\(_{19}C_9 = 92378\)

1. Calculate \(_{20}C_{10}\):

\(_{20}C_{10} = \frac{20 \times 19 \times 18 \times 17 \times 16 \times 15 \times 14 \times 13 \times 12 \times 11}{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}\) simplifies to calculate as:

\(_{20}C_{10} = 184756\)

1. Calculate \(_{21}C_{10}\):

\(_{21}C_{10} = \frac{21 \times 20 \times 19 \times 18 \times 17 \times 16 \times 15 \times 14 \times 13 \times 12}{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}\) simplifies to calculate as:

\(_{21}C_{10} = 352716\)

We are provided with the equation: \(\frac{a}{11} = \frac{b}{22} = \frac{c}{42}\). let's verify:

Substituting the calculated values of \(a\), \(b\), \(c\):

\(\frac{92378}{11} = 8398\)

\(\frac{184756}{22} = 8398\)

\(\frac{352716}{42} = 8398\)

Since all ratios are equal, the correct answer is:

\(\frac{a}{11} = \frac{b}{22} = \frac{c}{42}\).