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A $1 \times 1$ chessboard has one square, a $2 \times 2$ chessboard has five squares. Continuing along this fashion, what is the number of squares on the regular $8 \times 8$ chessboard?

  1. $64$
  2. $65$
  3. $204$
  4. $144$
  5. $256$
in Combinatory by Boss (30.8k points)
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0
I can solve this que with  the help of present of mind but why this que is the part of Permutations and combinations, where we are arranging and selecting.

6 Answers

+24 votes
Best answer

Now lets see how many $7 \times 7$ squares are possible

 

These two patterns can shift to right as well as follows:

So, $7\times 7$ squares possible is $4$

Now lets see how many $6 \times 6$ squares are possible

 

So, $6\times 6$ squares possible is $9$

Now lets see how many $5 \times 5$ squares are possible:

$4$ vertical moves $\times  4$ horizontal moves $=4^2$ possibilities.

Proceeding like this,

  • $8\times 8$  squares possible $: 1\times 1=1$
  • $7\times 7$  squares possible $:2\times 2=4$
  • $6\times 6$  squares possible $:3\times 3=9$
  • $5\times 5$  squares possible $:4\times4=16$
  • $4\times 4$  squares possible $: 5\times5=25$
  • $3\times 3$  squares possible $: 6\times6 =36$
  • $2\times 2$  squares possible $: 7\times7 =49$
  • $1\times 1$  squares possible $:8\times8=64$

Total squares $\quad: 204$

Now we can generalize like with $n \times n$ chess board total squares $=1^2+2^2+3^2+\ldots +n^2 = \frac{n(n+1)(2n+1)}{6}$

Correct Answer: $C$

by Veteran (119k points)
edited by
0
I could not understand your method. Can u please explain a bit more
0
Can u tell me how $2\times 2$ chess board 5 square possible?
+2
This is trivial.. I can count it by looking on it.. 4 squares of size 1*1. One big square of size 4*4
0
yes

but it is not $4\times 4$

it is $2\times 2$

right?
+1
Yes there are 4 squares in total in a grid of 2*2
0
Similarly I have done

In $8\times 8$ chess board $1\times 1$ square possible 64, i.e. $8\times 8$

                                        $2\times 2$    square possible  $7\times 7$

 

Say if  in a chess board we give number of each point like this

 

1              2                3               4................................

 

2

 

3

 

4

.

.

then $2\times 2$ square go like this

1 to 3 is one square, 2 to 4 next square like this.

like this u go and get $7\times 7$ squares for $2\times 2$ matrix.

got it?
0
Ok.. Got it... Thanks for detailed explanation..

I have few more queries in combinatorics... Can I ask u by sending message
+15 votes
No. of squares on chessboard of $n\times n$ is equal to sum of squares of $n$ terms for $8\times 8$ chessboard,

$\begin{align} &=\frac{n\left(n+1\right)\left(2n+1\right)}{6} \\&=\frac{8\times 9\times 17}{6}\\&=204.\end{align}$
by Boss (31.4k points)
edited by
0
Can u please explain why it worked this way
+3 votes

Let T(n,n) be the number of squares in a n*n chess board.

T(1,1) = 1 

T(2,2) (i.e) number of squares in a 2*2 chessboard = number of squares in a 1*1 chessboard + 22

Similarly T(n,n) = T(n-1,n-1) + n2  if n>2 with base conditions T(1,1) = 1 and T(2,2) = 5

Solving, we will get T(8,8) = 204.

                             

by Loyal (8k points)
edited by
0
After solving the relation -:

T(n,n) = $n*(n+1)*(2n+1)/6$
+1 vote
By simple observation it will be-

1^2 + 2^2+ 3^2+ ...................+ n^2.

where n is the size given, in the question its 8.

sum of squares of numbers.
by Boss (17.7k points)
+2

Yeah!.

 # squres in 8 ⨉ 8 chessboard = 82 (1 ⨉1 squres) +72 (2⨉2 squares)+62 (3⨉3 squares)+52 (4⨉4 squares)

+ 42 (5⨉5 squares)+32 (6 ⨉6 squares)+22 (7⨉7 squares)+12 (8 ⨉8 squares) = 204

0 votes

An $n \times n$ chessboard is made up of $n$ adjacent units on either side. Here, unit means side of each small square.

Now, $n$ adjacent units on horizontal side and $n$ adjacent units on vertical side make a square of $n \times n$ dimension.

Note: The horizontal and vertical sides must have common originating point.

 

Therefore, for a $n x n$ chessboard, following observations can be made-

Dimension i No. of Horizontal Sides of i Adjacent Units No. of Vertical Sides of i Adjacent Units Total No. of Squares
$n \times n$ $n - n + 1 = 1$ $n - n + 1 = 1$ $1^{2}$
$(n-1) \times (n-1)$ $n - (n-1) + 1 = 2$ $n - (n-1) + 1 = 2$ $2^{2}$
$(n-2) \times (n-2)$ $n - (n-2) + 1 = 3$ $n - (n-2) + 1 = 3$ $3^{2}$
... ... ... ...
$3 \times 3$ $n - 3+ 1 = (n-2)$ $n - 3+ 1 = (n-2)$ $(n-2)^{2}$
$2 \times 2$ $n - 2 + 1 = (n-1)$ $n - 2 + 1 = (n-1)$ $(n-1)^{2}$
$1 \times 1$ $n - 1 + 1 = n$ $n - 1 + 1 = n$ $n^{2}$

Therefore, total no. of squares = $1^{2} + 2^{2} + 3^{2} + ... + (n-2)^{2} + (n-1)^{2} + n^{2} = \frac{(n)(n+1)(2n+1)}{6}$  

Therefore,  total no. of squares in 8 x 8 chessboard = $\frac{(8)(8+1)((2 \times 8)+1)}{6} = 204$ 

by Junior (839 points)
edited by
0 votes
It can be solved by
1^2 + 2^2 + ... + 8^2 = (8 * 9 * 17) / 6 = 204
by (11 points)
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