Total number of ways of filling the 5 boxes numbered as (1, 2, 3, 4, and 5) with either blue or red balls 25 = 32.
Two adjacent boxes with blue can be obtained in 4 ways, i.e., (12), (23), (34) and (45).
Three adjacent boxes with blue can be obtained in 3 ways, i.e., (123), (234)and (345). Four boxes with blue can be obtained in 2 ways, i.e., (1234) and (2345). And five boxes with blue can be got in 1 way. Hence, the number of ways of filling the boxes such that adjacent boxes have blue
= (4 + 3 + 2 + 1) = 10.
Hence, the number or ways of filling up the boxes such that no two adjacent boxes have blue = 32 - 10 = 22.
22020 ÷ 0.011 = 2001818.181 ? 2000000
In 12 h, they are at a right angles, 22 times.
So, in 24 h, they are at right angles, 44 times.
P.W. = | 100 x T.D. | = | 100 x 168 | = 600. |
R x T | 14 x 2 |
∴ Sum = (P.W. + T.D.) = Rs. (600 + 168) = Rs. 768.
Each number is double the preceding one plus 1. So, the next number is (255 x 2) + 1 = 511.
Here, n(5) = {a, e, i,o, u}
and E = Event of selecting the vowel i = {i}
? P(E)= n(E)/n(S) = 1/5
? 90A/100 = 30B/100 = (30/100) x AC/100
? C = 100 x (100/30) x (90/100) = 300
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