1%=50
12%=600
B
390
12 %
5436
17,19,23,29
Let A, B and C be the three 6-faced dice.
Then, according to the question,
Since two dices has to be equal, that value can be any of the 6 faces, i.e., 6C1 cases.
Now for each case, 2 equal dices can be selected from 3 dices in 3C2 i.e., 3 ways.
And for each of the above, the third dice can have any of the 5 remaining faces
The possible outcomes are P(A)=61,P(B)=61,P(C)=65,P(A)=61,P(B)=65,P(C)=61 and P(A)=65,P(B)=61,P(C)=61
Hence the required probability = 61×61×65×6×3=21690=125
In the given series 5,6,7,8,10,11,14…
There are two series
First is 5,7,10,14…
Second is 6,8,11….
First series
5+2=7
7+3=10
10+4=14
Second series
6+2=8
8+3=11
11+4=15
Hence complete series is 5,6,7,8,10,11,14,15…
Sum of all numbers in series= 870 (I.e 312+162+132+142+122)
Average=sum of numbers/count of numbers
Average= 870/5
= 174
suppose
pipe:
A -30 hours A’s effeciency (60/30) =2
60( lcm of 30 and 20)
B- 20 hours B’s effeciency (60/20)=3
time taken by both to fill = 60/5 =12 as given in question (effeciencies of both a+b =2+3=5)
time taken by faster pipe i.e b = 60/3 =20
2, 6, 12, 20, 30, 42, 56, (…..)
Difference between 2 and 6 is = 4
Difference between 6 and 12 is = 6
Difference between 12 and 20 is = 8
Difference between 20 and 30 is = 10
Difference between 30 and 42 is = 12
Difference between 42 and 56 is = 14
So ne number will be with 16 Difference i .e 72
Therefore Answer will be 72
All books can be arranged in 10! ways. A single pair of books can be taken as a unit and arranged among the 8 others in 9! ways. The pair of books can also be interchanged and therefore rearranged in 2! ways. Thus the probability of the pair always being together is (9!*2!)/10!
10000
9/32