7474
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2011-02-01T00:00:01
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<div>Can you use this diagram to prove that the number of different
pairs of objects which can be chosen from six objects, $^6C_2$, is
$$1 + 2 + 3 + 4 + 5?$$</div>
<div>Generalise this to show that the number of ways of choosing
pairs from $n$ objects is</div>
<div> $$^nC_2 = 1 + 2 + ...+ (n-1) = \frac{1}{2}n (n -
1).$$ </div>
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<div class="framework">
<span style="font-style: italic;">Did you
know ... ?</span>
<br />
The sum of the first $n$ whole numbers is called a triangle number
because this sum can be represented geometrically by
a triangular array of dots. The sum is easily found by working
out the number of dots in the parallelogram formed by putting
two triangular arrays side by side.<br />
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The diagram shows that every combination of two elements chosen
from $6$ on the line $n = 6$ corresponds to exactly one dot in the
triangular array above. This triangular array above the
dotted line in the diagram represents the triangle number $T_5$.
Conversely the diagram also shows that every dot in the triangular
array for $T_5$ corresponds to one and only one of the choices of
pairs of elements in the line $n=6$ . Hence the number of
combinations of two elements chosen from $6$ is equal to $1 + 2 + 3
+ 4 + 5$.<br></br>
<br></br>
This generalises directly to finding the number of distinct pairs
of elements chosen from $n$ elements. Draw the triangular
array for $T_n$, that is rows of dots for $n = 1, 2, 3, ... n$. Now
in the same way as for $n=6$, every pair of elements chosen from
the bottom row can be joined by lines in the diagram to meet
in a single dot in the array for $T_{n-1}$. Conversely each
dot in $T_{n-1}$ corresponds to one and only one pair chosen from
$n$ elements on the bottom line. This shows that the number of ways
of choosing pairs from $n$ objects, generally denoted by
$^nC_2$, is $$T_{n-1} = 1 + 2 + ...+ (n-1).$$ Putting two $T_{n-1}$
triangular arrays side by side gives $n(n-1)$ dots so $$^nC_2
= T_{n-1} = \frac{1}{2}n (n - 1).$$<br></br>
<br></br>
<br></br>
<br></br>
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<?xml version="1.0" encoding="UTF-8"?>
<mdoxml version="1.0"><br></br>
The diagram shows that every combination of two elements chosen
from $6$ on the line $n = 6$ corresponds to exactly one dot in the
triangular array above. This triangular array above the
dotted line in the diagram represents the triangle number $T_5$.
Conversely the diagram also shows that every dot in the triangular
array for $T_5$ corresponds to one and only one of the choices of
pairs of elements in the line $n=6$ . Hence the number of
combinations of two elements chosen from $6$ is equal to $1 + 2 + 3
+ 4 + 5$.<br></br>
<br></br>
This generalises directly to finding the number of distinct pairs
of elements chosen from $n$ elements. Draw the triangular
array for $T_n$, that is rows of dots for $n = 1, 2, 3, ... n$. Now
in the same way as for $n=6$, every pair of elements chosen from
the bottom row can be joined by lines in the diagram to meet
in a single dot in the array for $T_{n-1}$. Conversely each
dot in $T_{n-1}$ corresponds to one and only one pair chosen from
$n$ elements on the bottom line. This shows that the number of ways
of choosing pairs from $n$ objects, generally denoted by
$^nC_2$, is $$T_{n-1} = 1 + 2 + ...+ (n-1).$$ Putting two $T_{n-1}$
triangular arrays side by side gives $n(n-1)$ dots so $$^nC_2
= T_{n-1} = \frac{1}{2}n (n - 1).$$<br></br>
<br></br>
<br></br>
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Weekly challenge 22: Combinations of Two
A weekly challenge: these are shorter problems aimed at Post-16 students or enthusiastic younger students.
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