Use the identity $(1+x)^{2n} = (1+x)^{n}(1+x)^{n}$ to show that $\sum_{k=0}^{n} \binom{n}{k}^{2} = \binom{2n}{n}^{2}$, where $n$ is a positive integer. A club has $2n$ members, with $n$ women and $n$ men. A group consisting of an even number $(0, 2, 4, \ldots, 2n)$ of members is chosen, with the number of men equal to the number of women. Show, giving reasons, that the number of ways to do this is $\binom{2n}{n}$. From the group chosen in part (ii), one of the men and one of the women are selected as leaders. Show, giving reasons, that the number of ways to choose the even number of people and then the leaders is $n^{2} \binom{n}{2} + 2^{2} \binom{n}{2} + \ldots + n^{2} \binom{n}{n}$. The process is now reversed so that the leaders, one man and one woman, are chosen first. The rest of the group is then selected, still made up of an equal number of women and men. By considering this reversed process and using part (ii), find a simple expression for the sum in part (iii).

SSCE HSC Mathematics Extension 1 Absolute value functions: Use the identity $(1+x)^{2n} = (

Use the identity $(1+x)^{2n} = (1+x)^{n}(1+x)^{n}$ to show that $\sum_{k=0}^{n} \binom{n}{k}^{2} = \binom{2n}{n}^{2}$, where $n$ is a positive integer - HSC - SSCE Mathematics Extension 1 - Question 14 - 2020 - Paper 1

Question 14

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Use the identity $(1+x)^{2n} = (1+x)^{n}(1+x)^{n}$ to show that $\sum_{k=0}^{n} \binom{n}{k}^{2} = \binom{2n}{n}^{2}$, where $n$ is a positive integer. A club has ... show full transcript

Worked Solution & Example Answer:Use the identity $(1+x)^{2n} = (1+x)^{n}(1+x)^{n}$ to show that $\sum_{k=0}^{n} \binom{n}{k}^{2} = \binom{2n}{n}^{2}$, where $n$ is a positive integer - HSC - SSCE Mathematics Extension 1 - Question 14 - 2020 - Paper 1

Step 1

Use the identity $(1+x)^{2n}$

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Answer

We start by applying the identity given in the question:

(1+x)^{2n} = (1+x)^{n} (1+x)^{n}.

By expanding each term in the product, we can express it as a sum of combinations:

(1+x)^{n} = \sum_{k=0}^{n} \binom{n}{k} x^{k}.

Therefore,

(1+x)^{2n} = \left( \sum_{k=0}^{n} \binom{n}{k} x^{k} \right) \left( \sum_{j=0}^{n} \binom{n}{j} x^{j} \right).

When we multiply these two sums, the coefficient of $x^{n}$ will correspond to selecting the same number of women and men, hence:

\sum_{k=0}^{n} \binom{n}{k}^{2} = \binom{2n}{n}.

Step 2

A group consisting of an even number (0, 2, 4, …, 2n) of members is chosen

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Answer

To choose a group with an equal number of men and women, we can always select:

$k$ women from the $n$ available, which can be done in $\binom{n}{k}$ ways,
$k$ men from the $n$ available, also done in $\binom{n}{k}$ ways.

The total number of ways to have $k$ men and $k$ women is:

\binom{n}{0} \binom{n}{0} + \binom{n}{1} \binom{n}{1} + \ldots + \binom{n}{n} \binom{n}{n} = \sum_{k=0}^{n} (\binom{n}{k})^{2} = \binom{2n}{n}.

Step 3

From the group chosen in part (ii), one of the men and one of the women are selected as leaders

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Answer

We can choose 1 woman and 1 leader in:

\binom{n}{1} \cdot n

for men. Given that we have chosen an even number, we have:

If we choose $k$ pairs ( $k$ women + $k$ men),
The total combinations will involve choosing leaders:

(k) \cdot (k) ext{ for both men and women.}

Thus the sum for leaders becomes:

Step 4

The process is now reversed

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Answer

In reversing the process, we select one man and one woman as leaders,

Then the remaining selections reduce the problem to selecting groups similar to part (ii):

Thus, we can write:

\left( n - 1 \right) \cdot \left( n-1 \right) ext{ for } (n-1) ext{ remaining pairs}.

So the total sum, combining all cases, gives:

$\sum_{k=1}^{n} n^{2} \cdot \binom{n-1}{k} \binom{n-1}{k-1} = (n-1) \cdot n.$

Use the identity $(1+x)^{2n} = (1+x)^{n}(1+x)^{n}$ to show that $\sum_{k=0}^{n} \binom{n}{k}^{2} = \binom{2n}{n}^{2}$, where $n$ is a positive integer - HSC - SSCE Mathematics Extension 1 - Question 14 - 2020 - Paper 1

Worked Solution & Example Answer:Use the identity $(1+x)^{2n} = (1+x)^{n}(1+x)^{n}$ to show that $\sum_{k=0}^{n} \binom{n}{k}^{2} = \binom{2n}{n}^{2}$, where $n$ is a positive integer - HSC - SSCE Mathematics Extension 1 - Question 14 - 2020 - Paper 1

Use the identity $(1+x)^{2n}$

A group consisting of an even number (0, 2, 4, …, 2n) of members is chosen

From the group chosen in part (ii), one of the men and one of the women are selected as leaders

The process is now reversed

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