search
Log In
Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true
17 votes
1.4k views

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
retagged by
1.4k views
1
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.

7 Answers

26 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: $C$


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
0

shortcuts:

total no of square possible in n *n chess board= (n(n+1)(2n+1))/6

total no of rectangles possible in n *n chess board= (n(n+1)/2)^2

17 votes
No. of squares on chessboard of $n\times n$ is equal to sum of squares of $n$ terms for $8\times 8$ chessboard,

$\begin{align} &=\frac{n\left(n+1\right)\left(2n+1\right)}{6} \\&=\frac{8\times 9\times 17}{6}\\&=204.\end{align}$

edited by
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.

                             


edited by
0
After solving the relation -:

T(n,n) = $n*(n+1)*(2n+1)/6$
0
Can u please explain what had insisted you to add "n^2" term to the recurrence relation ?
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.
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

1 vote

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 = $\frac{(8)(8+1)((2 \times 8)+1)}{6} = 204$ 


edited by
0 votes
It can be solved by
1^2 + 2^2 + ... + 8^2 = (8 * 9 * 17) / 6 = 204
0 votes
Ans : C

This can be solved by recognizing the pattern.

Let f(n) = Number of squares in n x n chessboard.

By drawing the chessboards and calculating number of squares manually for n=1,2,3,4 we get:-

f(1) =  1

f(2) = 4 + 1

f(3) = 9 + 4 + 1

f(4) = 16 + 9 + 4 + 1

These four observations are enough to conclude that f(n) = 1^2 + 2^2 + 3^2 +....+ n^2 = n(n+1)(2n+1)/6

Hence, f(8) = 204
Answer:

Related questions

26 votes
3 answers
1
1.9k views
There is a set of $2n$ people: $n$ male and $n$ female. A good party is one with equal number of males and females (including the one where none are invited). The total number of good parties is. $2^{n}$ $n^{2}$ $\binom{n}{⌊n/2⌋}^{2}$ $\binom{2n}{n}$ None of the above.
asked Dec 5, 2015 in Combinatory makhdoom ghaya 1.9k views
36 votes
3 answers
2
2.1k views
First, consider the tree on the left. On the right, the nine nodes of the tree have been assigned numbers from the set $\left\{1, 2,\ldots,9\right\}$ so that for every node, the numbers in its left subtree and right subtree lie in disjoint intervals (that is, all numbers in one subtree are less than all numbers in ... $2^{4}.3^{2}.5.9=6480$ $2^{3}.3.5.9=1080$ $2^{4}=16$ $2^{3}.3^{3}=216$
asked Dec 7, 2015 in DS makhdoom ghaya 2.1k views
18 votes
4 answers
3
1k views
Let $a, b, c$ be regular expressions. Which of the following identities is correct? $(a + b)^{*} = a^{*}b^{*}$ $a(b + c) = ab + c$ $(a + b)^{*} = a^{*} + b^{*}$ $(ab + a)^{*}a = a(ba + a)^{*}$ None of the above.
asked Dec 7, 2015 in Theory of Computation makhdoom ghaya 1k views
5 votes
2 answers
4
417 views
Let $A$ and $B$ be non-empty disjoint sets of real numbers. Suppose that the average of the numbers in the first set is $\mu_{A}$ and the average of the numbers in the second set is $\mu_{B}$; let the corresponding variances be $v_{A}$ and $v_{B}$ respectively. If the average of the elements in $A \cup B$ ... $p.v_{A}+ (1 - p). v_{B} + (\mu_{A}- \mu_{B})^{2}$
asked Dec 5, 2015 in Quantitative Aptitude makhdoom ghaya 417 views
...