30/10/2018 7:21 pm
PnC , Probability
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There are 40 lines in a plane of which a set of 12 lines are concurrent at A, another set of
15 lines are concurrent at B and the set of remaining lines are parallel. What is the number
of points of intersection of these 40 lines, given that the three sets are disjoint?
(A) 611 (B) 531 (C) 533 (D) 638
TG Team 30/10/2018 8:00 pm
Total number of points of intersection of 40 points : 40C2 = 780
Number of points of intersection when lines are parallel : 0
So remove 13C2 i.e 78 cases
12 lines are concurrent at A so remove 12C2 i.e 66 cases
15 lines are concurrent at B so remove 15C2 i.e 105 points
Hence , 780 - 105 - 78 - 66 + 2 = 533 intersection points .
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14/11/2018 11:23 pm
There are 33 similar chocolates to be distributed among 7 children such that each child gets distinct and natural number of chocolates. In how many ways it is possible ?
TG Team 14/11/2018 11:43 pm
first distribute 1, 2, 3,...., 7 chocolates
Remaining : 38 - 28 = 5 chocolates
now 5 chocalates can be distributed in
1 , 1, 1, 1, 1
1 , 1,1 2
1 , 1, 3
1, 4
1, 2, 2
2,3
5
7 ways
Further these 7 children can be arranged in 7! ways
Hence , 7 x 7! ways.
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15/11/2018 10:27 am
Sir , can you share the solution to this problem? I am getting B part as my answer , but in the key the ans given is C
This post was modified 6 years ago by Nikita Chawla
TG Team 15/11/2018 11:55 am
Hi Nikita
B is the correct answer.
Total number of triangles formed is nC3 , now consider the following cases –
1. Number of triangles having three sides common with the sides of polygon is 0.
2. Number of triangles having two sides common with the sides of polygon is n.
3. Number of triangles having one side common with the sides of the polygon = number of ways of selecting 3 points (vertices ) out of which only two are consecutive is n(n-4)C1 = n(n-4)
So number of required triangles is nC3 – n – n (n-4) = n ( n-4) (n-5)/6
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16/11/2018 11:52 am
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81 students appeared the examinations in physics, chemistry and biology. 46 students passed in physics, 47 students passed in chemistry, and 48 students passed in biology; 29 students passed in physics and chemistry, 27 students passed in chemistry and biology, and 13 students passed in physics and biology. The largest possible number that could have passed all the three examinations is:
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TG.Raman 16/11/2018 12:16 pm
E1+2E2+3E3 = 46+47+48
E1+E2+E3 = 81 (consider none fails for maximum value)
E2+3E3 = 29+ 27+ 13
Solve for E3 to get 9
Here E1 = number of students passed in exactly one exam
E2 = passed in exactly two exams
E3 = Number of students passed in all three exams.
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17/11/2018 3:18 pm
Is there any shortcut to find the number of squares in a rectangular chessboard? (eg. Number of squares in 8X10 chessboard?)
TG.Raman 17/11/2018 3:25 pm
Number_of_squares_and_rectangles_in a m x n matrix
Case 1 : m = n = k(say)
Squares = 1² + 2²+ .. + k² = k(k + 1)(2k + 1)/6
Rectangles = 1³+ 2³ + .. + k³ = [k(k + 1)/2]²
Case 2 : m ≠n
Squares = mn + (m - 1)(n - 1) + (m - 2)(n - 2) + .. + 0
Rectangles = (1 + 2 + 3 + .. + m)(1 + 2 + .. + n) = m(m + 1)n(n + 1)/4
TG.Raman 17/11/2018 3:28 pm
Hence, the number of squares in a 8 x 10 rectangular grid :
10 x 8 + 9 x 7 + 8 x 6 + 7 x 5 + 6 x 4 + 5 x 3 + 4 x 2 + 3 x 1 = 276 squares
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18/11/2018 2:16 pm
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18/11/2018 2:22 pm
3 points are randomly selected on a sphere,what is the probability that all of them lie in the same hemisphere?
Should it not be?-2πrsq./4πrsq.
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18/11/2018 2:25 pm
Sir,kindly post the Solution of both the questions
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14/12/2018 6:17 pm
An institute has 5 departments and each department has 50 students. If students are picked up randomly from all 5 departments to form a committee, what should be the minimum number of students in the committee so that at least one department should have representation of minimum 5 students?
1. 11
2. 15
3. 21
4. 41
5. None of the above
TG Team 14/12/2018 7:09 pm
The maximum number of students can be picked from each department such that 5 students are not selected from the same department is 4.
Therefore, after 4 students from each department are selected i.e 4 x 5 = 20 , the 21st student selected will be the fifth student to be selected from one of the 5 departments.
Hence, , 20+1 = 21 students should be selected in total to ensure that at least five students from one of the departments is selected.
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15/12/2018 12:29 pm
In how many ways can we select 3 numbers from first 20 positive integers, such that their product is always divisible by 5?
Sir, won't the answer be 580?
TG.Raman 15/12/2018 1:02 pm
yes 580 is the correct answer .
Approach : 1
select 3 numbers in 20C3 ways.
cases when none of the selected number is a multiple of 5 : 16C3 ways
20C3 - 16C3 = 580
Approach : 2
4C1 x 16C2 + 4C2 x 16C1 + 4C3 = 580
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08/07/2019 9:18 pm
How many 3-letter words can be formed using the letters of the word 'INSTITUTE'?
Gaurav Jain 12/07/2019 3:40 pm
case 1 - when all letters are distinct
then there are 6 different letters hence 6P3 = 120
Case 2 - when 1 letter repeated twice
then it can be either I or T means 2C1 cases and 1 more letter means 4C1 case
further arrangements 3! , hence total 2C1*4C1*3! = 48
Case 3 - when all letters are same , that is only one case TTT
hence total 169 cases
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19/07/2019 11:50 pm
In how many ways can you select exactly 7 letters from 3A, 4B, 2C and 1 D?
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19/07/2019 11:54 pm
x+ y+ z=19, x≤ y≤ z, find total number of whole number solutions?
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20/07/2019 12:00 am
Find the number of ways to distribute 2n different balls into
(a) n equal lots
(b) equally among n children
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17/08/2019 9:02 am
Hello, Please solve :
How many different size rectangles are there in a square grid of 5 x 7 ?