Problems based on grids/chessboards have been asked in the CAT and other MBA entrance exams over the years. They seem really difficult when you encounter them for the first time but once you get the hang of things they become really simple. The key lies in understanding the basic concepts involved.
The most common grid structure that we are all familiar with is the chessboard. Let us look at some of the common questions based upon grid.
Q1. What is the number of squares on a chessboard?
Squares of size 1x1 = 8^2 = 64
Squares of size 2x2 = 7^2 = 49
Squares of size 8x8 = 1^2 = 1
Total number of squares = 1 + 4 + 9 .. 49 + 64 = 204
You could have also used the formula 1^2 + 2^2 + 3^2 … n^2 = n(n+1)(2n+1)/6
In this case, you would have got = 8*9*17/6 = 12*17 = 204
Q2. What is the total number of rectangles on a chessboard?
To form a 8 x 8 chessboard, we need 9 horizontal and 9 vertical lines.
If we select any 2 lines from the 9 horizontal lines and any 2 lines from the 9 vertical lines, we will get a rectangle
Total number of rectangles = 9C2 * 9C2 = 36*36 = 1296
Q3. In how many ways can you place 2 rooks on a chessboard such that they are not in attacking positions?
The first rook can be placed in 64 ways
The second rook cannot be placed in the same row or the same column. So, it has 7 rows and 7 columns left for it. It can be placed in 49 ways.
But the order in which the rooks are placed is not important. So, it will be divided by 2!
Total ways = 64*49/2 = 1568
Q4. In how many ways can you place 8 rooks on a chessboard such that they are not in attacking positions?
The first rook can be placed in 64 ways.
The second rook can be placed in 49 ways.
The eigth rook can be placed in 1 way
But the order in which the rooks are placed is not important. So, it will be divided by 8!
Total ways = 64*49*36…4*1/8! = 8! = 40320
Q5. In how many ways is it possible to choose a white square and a black square on a chess board so that the squares must not lie in the same row or column? [CAT 2002]
The white square can be chosen in 32 ways.
If we remove the row and the column which contains the chosen white square, we will be left with 7 Rows & 7 Columns containings a total of 49 squares (24 black and 25 white). We would have removed 15 squares (7 white and 8 black)
Required ways = 32*24 = 768
Q6. In how many ways can you go from A to B? (Shortest path)
To go from A to B, you need to make 4 right moves (R) and 6 up moves(U).
One of the possible paths could be RRRRUUUUUU
Any rearrangement of the above would give you a different shortest path from A to B.
RRRRUUUUUU can be rearranged in 10!/(4!6!) = 210 ways because among the 10 entities four Rs are identical and 6 Us are identical.
Another way of solving this could be:
From the 10 moves that you have to make select the 4 moves which will be Right moves. This can be done in 10C4 = 10!/(4!6!) = 210 ways
Q7. In how many ways can you go from A to B? (Shortest path)
To go from A to B, you will have to follow the path APQB
A to P can be done in 4C2 = 6 ways (Select 2 right moves from a total of 4 moves)
P to Q is only one way (The path through the red rectangle)
Q to B can be done in 3C1 = 3 ways (Select 1 right move from a total of 3 moves)
Total ways = 6 x 1 x 3 = 18 ways
Q8. In how many ways can you go from A to B? (Shortest path)
Total ways (if the red rectangle was not there) = 10C4 = 210
PQ is the road which has been removed. It would make all the routes which included the road PQ as invalid.
So, all routes consisting of APQB will be invalid
A to P = 5C3 = 10
P to Q = 1
Q to B = 4C3 = 4
Invalid ways = 10 x 1 x 4 = 40
Total valid ways = 210 – 40 = 170 ways
[CAT 2008] The figure below shows the plan of a town. The streets are at right angles to each other. A rectangular park (P) is situated inside the town with a diagonal road running through it. There is also a prohibited region (D) in the town.
Q9. Neelam rides her bicycle from her house at A to her office at B, taking the shortest path. Then the number of possible shortest paths that she can choose is
Let us call the road through park as MN.
The path Neelam will take is AMNB.
Neelam can go from A to M in 4C2 = 6 ways.
M to N in 1 way
N to B in 6C2 = 15 ways.
Number of possible shortest paths = 6*15 = 90
Q10. Neelam rides her bicycle from her house at A to her club at C, via B taking the shortest path. Then the number of possible shortest paths that she can choose is
Neelam can go from A to B in 90 ways.
From B to C via N = 6 (and not via M)
From B to C via M = 7 ways
In all 90*(6+7) = 1170 ways
I hope you enjoyed this post and if you get a grid based question in CAT, you will be able to crack it.
Author Ravi Handa has taught Quantitative Aptitude at IMS for six years. An alumnus of IIT Kharagpur where he studied a dual-degree in computer science, he currently runs www.handakafunda.com.