1. Number of selections of atleast 10 articles from 19 different articles is.
A) 2^19 B) 2^18 C) 2^19 - 1 D) 2^18 - 1 E) None
please provide the solutions for questions 41,46-49 (with explanation).
47th
Number of ways of selecting 3 different letters:
=4C3=4 ways
Number of ways to select 2 similar and 1different letter:
=4C1×3C2=12 ways
Number of ways select 3 similar letter =2 ways
total=4+12+2=18 ways
in other questions also you have to make cases and then add them
41st
No of groups = 4
Towns in each group = 3
For towns belonging to the same group.
Lets say a1, a2, a3.
Each requires 3 connections with other town in same group.
a1 to a2 = 3 connections (same as a2 to a1)
a1 to a3 = 3 connections (same as a3 to a1)
a2 to a3 = 3 connections (same as a3 to a2)
We have total of 9 connections required in same group.
4 groups --> So total no of connections = 4*9 = 36
Now, for connections between the groups = a,b,c,d
a1 to (b1,b2,b3) = 3 connections
a1 to (c1,c2,c3) = 3 connections
a1 to (d1,d2,d3) = 3 connections
Total connections = 9
Similarly a2 and a3 to other groups will have 9 + 9 = 18 connections.
As explained above, From group 'a' to other groups b,c,d = 9*3 = 27 connections
From group 'b' to group c,d = 9*2 = 18 connections
From group 'c' to group d = 9*1 connections
So total no of connections required = 4*9 + 9*(3+2+1) = 90 connections.
solution for question 13
Ans 13
solution for question 15 please
here introduce the fifth variable to minimize the value of other four variables.
ans 15
solution for question 18 please
there is a general formula to solve this question
4n^2+2
therefore 4*15*15 + 2= 902
Three Cases
(1) none of them 0 .
x + y + z = 10 => `14C2
14C2 x 8 = 728 cases
(2) One of them 0.
3C2 x 14 = 42
42 x 4 = 168 cases
(3) Two 0.
3C1 x 1
3 x 2 = 6 cases
Total : 6 + 168 + 728 = 902 Solutions .
Some important results :
Total Number of ontegral solution :
| X | = n ; 2
| X | + |Y | = n ; 4n
|X | + |Y | + |Y | = ; 4n² + 2
|W | + |X| + |Y| + |Z| = n ; 8n/3 × (n² + 2)
solution for question 20 please
(a)
For 4 dices being thrown sum or we can say outcomes can be:
4,5,6,………………14……………..,22,23,24
We also know that these are equally spaced and aligned with symmetry.
Mid point is (24+4)/2 = 14. The left side of the mid point is mirror image of the right part.
No of ways to get 19=(14+5) = no of ways to get (14-5)= 9
9-1C4-1= 8C3=56 ways
same can be done with b part also
(6-a) + (6-b) + (6-c) + (6-d) = 19
a + b + c + d = 24 - 19 = 5
Hence , 8c3 = 56 ways .
____________________________________
(6-a) + (6-b) + (6-c) + (6-d) = 18
a + b + c + d = 6
Total : 9C3 solutions
Remove cases when a or b or c or d =6. ( 4 ways )
Hence , 80 ways ,
solution for question 35 please
solution for question 37 please.
8empty seats left (denoted by O)
like O O O O O O O O
there are 7+2=9 interspace to place the 6 groups,
this can be done in 9C6
if the persons' order is alterable,
then the answer is 9C6 * 12!
solution for question 38 please
consider all outcomes for sum to be 11 following cases are there
(6,4,1) =3! ways= 6 ways
(6,3,2)=3!=6
(5,5,1)=3!/2!=3
(5,4,2)=3!=6
(5,3,3)=3!/2!=3
(4,4,3)=3!/2!=3
total 27 ways
Alternatively :
( 6 -a ) + ( 6 - b) + ( 6 - c) = 11
a+ b + c = 7
9C2 = 36 solutions
Remove cases when a or b, or c greater than 5 .
a' + 6 + b + c = 7
a' + b + c = 1 => 3C2 i.e 3 solutions.
36 - 3 × 3 = 27 ways .
solution for question 40 please
Two cases possible :
(1) 3 , 3, 1 => 3C2 x 7C3 x 4C3 x 1 = 420 ways
(2) 3 , 2, 2 => 3C2 x 7C3 x 4C2 x 2C2 = 630 ways
Total : 420 + 630 = 1050 ways
Could you please share the solutions for the both the questions?
14.
E is four steps ahead of A. So it is not possible to reach E from A in (2n - 1) i.e. odd number of steps. Hence Option ( A)
15.
is it 33 ?
I am not sure of the answer. But could you please share you approach?
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
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 .
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 ?
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.
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
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
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: |
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.
Is there any shortcut to find the number of squares in a rectangular chessboard? (eg. Number of squares in 8X10 chessboard?)
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
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
