Finding a:

Case 1: All the boxes have 1 ball each => 7! ways
Case 2: One box has 0 balls and one box has 2 balls, all the other boxes have 1 ball each => 7C1 (for selecting the box with 0) * 6C1 (selecting the box with 1)* 7C2 (selecting 2 balls to be placed in the box) * 5! (distribution of the remaining balls in the 5 boxes) ways
Case 3: Case of (2 2 1 1 1 0 0) => 7C2 (for selecting 2 boxes that have 0 balls) * 5C2 (for selecting 2 boxes that will have 2 balls each)* 7C2 (balls to be placed in the first box)* 5C2 (balls to be placed in the second box) * 3! (distribution of the remaining 3 balls) ways
Case 4: Case of (2 2 2 1 0 0 0) => 7C3 (selecting 3 boxes that will have 0 balls each) * 4C3 (selecting the box with 1 ball) * 7C2 (selecting 2 balls) * 5C2 (selecting the next set of 2 balls) * 3C2 (selecting the third set of 2 balls) ways

Sum of these = 463680

Finding b: There in only 1 way in which 7 distinct balls can be put in 7 similar boxes such that each box has only 1 ball

So, a/b = 463680

x=1, z=2,y=3 ans is 108

x + y/3 + y/3 + y/3 + z/2 + z/2 = 6

AM >= GM

(x + y/3 + y/3 + y/3 + z/2 + z/2)/6 >= $$(x*y/3*y/3*y/3*z/2*z/2)^{1/6}$$

=> 1 >= $$(x*y/3*y/3*y/3*z/2*z/2)^{1/6}$$
=> 1 >= $$x*y^3*z^2 / (3^3*2^2)$$
=> $$x*y^3*z^2$$ <= 27*4 = 108

Max value = 108

There are three results - win, draw, loss

Select the 12 matches in $$^{15}C_{12}$$ ways. For these, the correct prediction is possible in only 1 way.
For each of the other three matches, a wrong prediction can be done in 2 ways. So, total number of ways of predicting wrong = $$2^3$$

Total number of ways = $$^{15}C_{12}$$*$$2^3$$ = 3640 ways

The maximum length of any side can be 49. (If one side is 50, then the sum of the other two sides, 49, will be less than this side)
Let the lengths of the three sides be 49-x, 49-y, 49-z, where x, y, z can vary from 0 to 48.
Now, since the perimeter is 99, 49 - x + 49 - y + 49 - z = 99 => x+y+z = 48 and 0 <= x, y, z <= 48.

This can be found out by using the number of integral solutions formula. $$^{n+r-1}C_{r-1}$$ = $$^{48+3-1}C_{3-1}$$ = $$^{50}C_2$$ = 49*25 =1225
This is the total number of triangles.

To get scalene triangles, we have to get the number of integral solutions for x+y+z = 48, where 0 <= x < y < z <= 48

If x = y = 0, there is one solution; similarly for x = y = 1 and so on till x = y = 24 -> 25 solutions in all (This also takes care of one equilateral triangle case).
Again, two variables can be equal in three ways - x,y ; y,z or z,x. So, the number of isosceles triangles is 24*3 = 72
Number of equilateral triangles = 1
So, number of scalene triangles = 1225 - 73 = 1152
But, there are for ordered x, y and z. Since we want the general case of unordered variables, we have to divide this by 3! = 6

So, number of scalene triangles =1152/6 = 192

Download Geometry formulas for CAT PDF

india have won the toss.

have

The ratio of speeds of the boy and the dog are needed to answer this question

f(x) = $$16^x/(16^x + 4)$$
f(1-x) = $$16^{1-x}/(16^{1-x}+4)$$ = $$16*16^x/[16^x*(16+4*16^x)]$$ = $$4/(16^x+4)$$
f(x) + f(1-x) = 1
f(1/852)+f(851/852) = 1
...
So, required sum = 1 + 1 + ... (425 times) + f(426/852) = 425 + f(1/2) = 425 + 0.5 = 425.5

Hi Shubham,

Sorry for the inconvenience and thanks for the feedback. We are working on it and it will be resolved very soon.

There are 5 flags.
Total Number of signals made using 5 different colored flags = 5p1+5p2+5p3+5p4+5p5 = 325

325=5P1+5P2.....+5P5

Mom did not start from home at 12 pm. She usually starts before 12pm so that she can reach the bus stop at 12pm. So the son only walks for 50 minutes and reaches the place from where his mom takes 10 minutes to reach the bus stop.