ONE MORE STEP TOWARD BETTER TOMORROW : CHAPTER 8

Tuesday, 5 August 2014

CHAPTER 8 - More Applications

By now, you should have a pretty good idea about the basics of permutations and combinations. In this section, we will encounter more advanced problems.

All the questions discussed in the following pages are directly or indirectly based on the concepts already discussed in the previous sections. In case you find anything confusing, refer to the relevant parts again.

Example: 1

On a standard $8 \times 8$ chessboard, find the

	(a) number of rectangles
	(b) number of squares

Solution: 1-(a)

Visualise any arbitrary rectangle on the chessboard, say the one depicted on the left in the figure below:

As explained in the figure above, any rectangle that we select can be uniquely determined by specifying the pair of lines $X$ and $Y$ that make up the vertical edges of the rectangle and the pair of lines $P$ and $Q$ that make up the horizontal edges of the rectangle.

On the chessboard, there are $9$ vertical lines available to us from which we have to select $2$ . This can be done in ${}^9{C_2}$ ways. Similarly, $2$ horizontal lines can be selected in ${}^9{C_2}$ ways. Thus, the total number of rectangles that can be formed is ${}^9{C_2} \times {}^9{C_2} = 1296$ .

Solution: 1-(b)

To select a square, observe that the pair of lines $X$ and $Y$ must have the same spacing within $X$ and $Y$ as the pair of lines $P$ and $Q$ . Only then can the horizontal and vertical edges of the selected rectangle be of equal length (and thus, the selected rectangle is actually a square).

In how many ways can we select a pair $(X, Y)$ of lines which are spaced a unit distance apart ? Its obviously $8$ . Corresponding to each of these $8$ pairs, we can select a pair $(P, Q)$ of lines in $8$ ways such that $P$ and $Q$ are a unit distance apart. Thus, the total number of unit squares is $8 \times 8 = 64$ (This is obvious otherwise also). Now we count the number of $2 \times 2$ squares. In how many ways can we select a pair $(X, Y)$ of lines which are $2$ units apart ? A little thought shows that it will be $7$ . Corresponding to each of these $7$ pairs, we can select a pair $(P, Q)$ of lines in $7$ ways such that $P$ and $Q$ are $2$ units apart. Thus, the total number of $2 \times 2$ squares is $7 \times 7 = 49$ .

Reasoning this way, we find that the total number of $3 \times 3$ squares will be $6 \times 6 = 36$ , the total number of $4 \times 4$ squares will be $5 \times 5 = 25$ and so on.

Thus, the total number of all possible squares is

$\mathop {64}\limits_{\scriptstyle\,\,\,\,\,\,\,\,\,\,\,\, \uparrow \hfill\atop \scriptstyle\,1 \times 1\,\,{\rm{squares}}\hfill} + \mathop {49}\limits_{\scriptstyle\,\,\,\,\,\,\, \uparrow \hfill\atop {\scriptstyle\,\,\,\,2 \times 2\hfill\atop \scriptstyle{\rm{squares}}\hfill}} + \mathop {36}\limits_{\scriptstyle\,\,\,\,\,\, \uparrow \hfill\atop {\scriptstyle\,\,\,3 \times 3\hfill\atop \scriptstyle{\rm{squares}}\hfill}} + \ldots + \mathop 4\limits_{\scriptstyle\,\,\,\,\,\,\, \uparrow \hfill\atop {\scriptstyle\,\,\,\,7 \times 7\hfill\atop \scriptstyle{\rm{squares}}\hfill}} + \mathop 1\limits_{\scriptstyle\,\,\,\,\,\,\, \uparrow \hfill\atop {\scriptstyle\,\,\,\,8 \times 8\hfill\atop \scriptstyle{\rm{squares}}\hfill}} = 204$

Example: 2

Consider an $n$ -sided convex polygon.

	(a) In how many ways can a quadrilateral be formed by joining the vertices of the polygon ?
	(b) How many diagonals can be formed in the polygon?

Solution: 2-(a)

Observe that a selection of any $4$ points out of the $n$ vertices of the quadrilateral will give rise to a unique quadrilateral (since the polygon is convex, the problem of our selection containing all or $3$ collinear points does not exist). $4$ points out of $n$ can be selected in ${}^n{C_4}$ ways. Thus, we can have ${}^n{C_4}$ different quadrilaterals.

Solution: 2-(b)

To form a diagonal, we need $2$ non-adjacent vertices ( because $2$ adjacent vertices will form a side of the polygon and not a diagonal). The total number of ways of selecting $2$ vertices out if $n$ is ${}^n{C_2}$ . This number also contains the selections where the $2$ vertices are adjacent. Those selections are simply $n$ in number because the polygon has $n$ sides. Thus, the total number of diagonals is

	${}^n{C_2} - n$
	$= \dfrac{{n(n - 1)}}{2} - n$
	$= \dfrac{{n(n - 3)}}{2}$

Tuesday, 5 August 2014

CHAPTER 8 - More Applications

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