Permutation and combination involving game of 64 Squares
Question 1:Find the number of squares which can be formed from 8 cm x 8 cm chessboard?
Solution:
Figure ALet us consider a 8cmX8cmchessboard as shown in the
in which AB=BC=………=HI=1 cm
Let us calculate those squares which have area 1 cm2.
We have 8 such squares in each row and column
Therefore in totality we have 8×8 = 82 squares
Now we calculate those squares which have area 4 cm2
i.e. we pick those squares having side length 2. Obvious choices are picking squares with sides AC ,BD,CF,DF,EG,FH,GI i.e. 7 possibilities in a row.Similalrly 7 choices (AK,JL,KM,LN,MO,NP,OQ)possible as we move down in a column i.e sides of length 2cm .So 7 squares in each row and column therefore in totality we have total of 7×7 = 72 squares
Next we calculate those squares which have area 9 cm2
i.e. we pick those squares having side length 3. Obvious choices are picking squares with sides AD ,BE,CF,DG,EH,FI i.e. 6 possibilities in a row.Similalrly 6 choices (AL,JM,KN,LO,MP,NQ)possible as we move down in a column i.e. sides of length 3cm .So 6 squares in each row and column therefore in totality we have total of 6×6 = 62 squares .
Similarly, number of squares having side length 4 will be 52 squares
Similarly, number of squares having side length 5 will be 42 squares
Similarly, number of squares having side length 6 will be 32 squares
Similarly, number of squares having side length 7 will be 22 squares
Similarly, number of squares having side length 8 will be 12 squares
GENERALISATION: Number of squares which can be formed from N x N chessboard
Question 2:Find the number of rectangles which can be formed from 8 cm x 8 cm chessboard?
Solution:
METHOD 1
We first calculate rectangles of size 8×8 clearly it will be only 1 (Whole of chess board)
Now we calculate rectangles of size 1×2
Let us understand this in detail ,if we fix one side as AJ (1 cm) .Then possibilities of other side of rectangles are AC ,BD,CF,DF,EG,FH,GI i.e. 7 possibilities .
Similarly for JK,KL,…………..,PQ we have 7 possibilities each.Therefore in totality 8 x7 rectangles of size
-1×2 are possible
Analogous to this we can directly say rectangles of size 2×1 will be 7×8
Now we calculate rectangles of size 1×3
Let us understand this in detail ,if we fix one side as AJ (1 cm) .Then possibilities of other side of rectangles are AD ,BE,CF,DG,EH,FI i.e. 6 possibilities .
Similarly for JK,KL,…………..,PQ we have 6 possibilities each.Therefore in totality 8 x6 rectangles of size
1×3 are possible
Analogous to this we can directly say rectangles of size 3×1 will be 6×8
Similarly rectangles of size 1×4 will be 8×5 and rectangles of size 4×1 will be 5×8 and so on
Hence total number of rectangles will be
METHOD 2
Number of selections of 2 consecutive lines out of the given 8 parallel lines =8
Number of selections of 3 consecutive lines out of the given 8 parallel lines =7
Number of selections of 4 consecutive lines out of the given 8 parallel lines =6
Number of selections of 5 consecutive lines out of the given 8 parallel lines =5
…………………………………………………………………………………………………………………………
Number of selections of 8 consecutive lines out of the given 8 parallel lines =1
Therefore number of rectangles = (1+2+3+….+8) (1+2+3+….+8)
=1296
GENERALISATION:
Number of rectangles which can be formed from N x N chessboard
Question 3:
Find the number of pure rectangles which can be formed from 8 cm x 8 cm chessboard?
Solution:
We calculated earlier, Number of rectangles formed in 8 cm x 8 cm chessboard =1296
Number of squares in 8 cm x 8 cm chessboard =204
Thus number of pure rectangles = No of rectangles – No of Squares=1296 – 204=1092
The next obvious generalization which can be done on the basis of what we learnt above is
GENERALISATION:
Solution:
We know parallelogram is such a figure in which opposite sides are parallel. Out of the given 9 parallel HORIZONTAL lines we can choose any two lines which can be done inways. Similarly
Out of the given 9 VERTICAL set of parallel lines we can choose any two lines which can be done in
Question 5: Given is a road map of a city in form of chess board (as shown in figure A). Find the number of different paths that a person can take to reach from one corner to diagonally opposite corner of the city (From corner Q to I)?
Solution:
Let us take simplest path in going from to Q to I by taking straight road from Q to A then straight road from A to I i.e. he goes from
QP -> PO ->ON->NM->ML->LK->KJ->JA->AB -> BC ->CD->DE->EF->FG->GH->HI
Notice that he took 8 vertical steps (VVVVVVVV where V denotes vertical step) and 8 horizontal steps(HHHHHHHH where H denotes horizontal steps) i.e. he moved in following manner
VVVVVVVVHHHHHHHH
Or even if person chooses some other path as well then also it will involve 8 vertical steps or 8 horizontal steps arranged in some other pattern
We know the no of ways of arranging the word with 8 V and 8 H are
Question 6: Given is a road map of a city in form of chess board (as shown in figure B where a portion in between is darkened to depict that in between roads are under construction). Find the number of different paths that a person can take to reach from A to X first then from Y to B under the condition that a person cannot take in between roads which are under construction?
Solution:
Let us calculate the number of ways a which a person can go from A to X
In going from A to X a person will take 3 horizontal
steps (HHH where H denotes horizontal steps)and 3 vertical
VVV where V denotes vertical step) i.e. he moved in
following manner
VVVHHH
We know the no of ways of arranging the word with 3 V and 3 H are
Now as per given in between roads are under construction so he cannot take in between roads which are under construction so there are only 2 ways to go from X to Y i.e. from X ->P->Y or X ->Q->Y
In going from Y to B a person will take 2 horizontal
Steps (HH where H denotes horizontal steps)and 2 vertical( VV where V denotes vertical step) i.e. he moved in following manner VVHH.
We know the no of ways of arranging the word with 2 V and 2 H are
Thus total number of ways in going from A to B = (20)(2)(6)=240 ways
Question 7: Two squares are chosen at random from small squares drawn on a chess board. Find the number of ways in which 2 squares can be chosen such that they have exactly one corner in common?
Solution:
Two squares selected can have a corner common if they are selected from two consecutive rows or columns. The number of ways to select two consecutive rows(or columns) are 7.
Two squares selected can have a corner common if they are selected from two consecutive rows or columns. The number of ways to select two consecutive rows(or columns) are 7.
For each pair of two consecutive rows(or columns) ,the number of pairs of squares having exactly one common corner =(2)(7)=14 ways
Total number of favourable selections= (14)(7)=98
Question 8: Find the number of ways in which a white and a black square on a chess board be chosen so that two squares do not belong to same row or column?
Solution:
In a standard chess board we have 32 white squares and 32 black squares .Thus any white square can be chosen in 32 ways and once a particular white square is chosen we cannot pick any other square row from that particular row or column(i.e. in that row or column we will have 4 black and 4 white squares)Hence we are left with 32 – 8 = 24 squares .Therefore , the number of ways in which a white and a black square on a chess board be chosen so that two squares do not belong to same row or column are (32)(24)=768 ways
Question 9: In the given Grid(A Chessboard designed in such a way that middle square of outer chess board 8×8 contains 1 more smaller grid of dimension 4×4 ), Find the number of squares of all possible dimensions?
Solution:
Clearly the number of squares in the outer 8×8 chess board = 82+72+62+52+42+32+22+12 =208
Clearly the number of squares in the inner 4 x 4 grid = 42+32+22+12=30
But one square is common. Thus required number of squares =208+30 -1 =237
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