Greatest Value of 2nCr

The question asks us to determine the relationship between the greatest values of two combination expressions: $2nCr$ and $(2n-1)Cr$. The value of a combination, $\text{nCr}$ (read as 'n choose r'), typically reaches its maximum when $r$ is close to $n/2$.

For the expression $2nCr$, the total number of items is $2n$. The greatest value occurs when $r$ is exactly half of the total, which means $r = \frac{2n}{2} = n$.

Therefore, the greatest value, denoted by $a$, is $2nCn$. Using the standard formula for combinations, $\text{nCr} = \frac{n!}{r!(n-r)!}$, we can express $a$ as:

$$a = 2nCn = \frac{(2n)!}{n!(2n-n)!} = \frac{(2n)!}{n!n!}$$

Greatest Value of 2n-1Cr

For the expression $(2n-1)Cr$, the total number of items is $2n-1$. The greatest value occurs when $r$ is closest to $\frac{2n-1}{2}$.

Calculating $\frac{2n-1}{2}$ gives $n - 0.5$. Since $r$ must be an integer, the integers closest to $n - 0.5$ are $n-1$ and $n$. The combination value is the same for both $r=n-1$ and $r=n$, thanks to the identity $\text{nCr} = \text{nC(n-r)}$.

Let's calculate this greatest value, denoted by $b$, using $r = n-1$:

$$b = (2n-1)C(n-1) = \frac{(2n-1)!}{(n-1)!((2n-1)-(n-1))!}$$

Simplifying the denominator's second term: $((2n-1)-(n-1))! = (2n-1-n+1)! = n!$. So,

$$b = \frac{(2n-1)!}{(n-1)!n!}$$

Relating Greatest Values: a and b

We have the expressions for $a$ and $b$:

• $a = \frac{(2n)!}{n!n!}$
• $b = \frac{(2n-1)!}{(n-1)!n!}$

To find the relationship between $a$ and $b$, let's rewrite $a$ by expanding the factorial $(2n)!$ and $n!$ in the denominator:

$$a = \frac{(2n)!}{n!n!} = \frac{2n \times (2n-1)!}{(n \times (n-1)!) \times n!}$$

Now, let's rearrange the terms to see if we can isolate the expression for $b$:

$$a = \left( \frac{2n}{n} \right) \times \frac{(2n-1)!}{(n-1)!n!}$$

We recognize that the second part of the expression, $\frac{(2n-1)!}{(n-1)!n!}$, is exactly $b$. Substituting $b$ into the equation:

$$a = \left( \frac{2n}{n} \right) \times b$$

Simplifying the term $\frac{2n}{n}$ gives $2$. Therefore, the relationship is:

$$a = 2b$$

Comparison of Combination Maximums

The calculation shows that the greatest value of $2nCr$, represented by $a$, is twice the greatest value of $(2n-1)Cr$, represented by $b$.

This confirms the relationship $a = 2b$. This result matches the first option provided in the question.