Given expression = log x + log1/x
= log x + log 1 - log x
= log 1
= 0
Required number of ways = (2 + 1) (3 + 1) (4 + 1) - 1 = 59
We have 9 letters 3 a's, 2 b's and 4 c's. These 9 letters can be arranged in 9! / (3! 2! 4!) = 1260 ways.
35Cn + 7 = 35C4n - 2
We know that if nCx = nCy then, Either x = y or x + y = n
Case 1: x + y = n
? n + 7 + 4n - 2 = 35
? 5n + 5 = 35
? 5n = 30
? n = 6
Case 2: x = y
? n + 7 = 4n - 2
? 4n - n = 2 + 7
? 3n = 9
? n = 3
We have If 2n + 1Pn - 1 / 2n - 1Pn = 3/5
? 5 x 2n + 1Pn - 1 = 3 x 2n - 1Pn
? {5 x (2n + 1)!} / {(n + 2 )!} = 3 x {(2n - 1)!} / {(n - 1)!}
? 10(2n + 1) = 3(n + 2) (n + 1)
? 3n2 - 11n - 4 = 0
?(3n + 1 ) (n - 4) = 0
? n = 4
The hundred place will be reserved for 3 or 2
5 digits are free to fill rest two places i,e., of tens and unit.
Number of required 3 digit number = 2 x 5C2 = 20
Required no. of ways = [4C1 x 6C3] + [4C2 x 6C2] + [4C3 x 6C1] + [4C4]
= 80 + 90 + 24 + 1 = 195.
If there were no three points collinear. We should have 10C2 lines but since 7 points are collinear we must subtract 7C2 lines and add the one corresponding to the line of collinearity of the seven points.
Thus, the required number of straight lines .
= 10C2 - 7C2 + 1 = 25
Let there be n persons in the room. The total number of hand shakes is same as the number of ways of selecting 2 out of n.
nC2 = 66
? n(n - 1) / 2! = 66
? n2 - n - 132 = 0
? (n - 12) (n + 11) = 0
? n = 12
Taping all vowels (IEO) as a single letter (since they come together ) there are six letters with two 'R' s
Hence no. of arrangement = 6!/2! x 3! = 2160
[3 vowels can be arranged in 3! ways among themselves, here multiplied with 3!. ]
Possible arrangements are 7!/2! 2! = 1260
[Division by 2 times 2! is because of the repletion of E and R .]
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