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tobeiitian,
T5 = 24 * T3
=> 11C4 * x^4 = 24 * 11C2 * x^2
=> 11!/4!7! * x^2 = 24 * 11!/2!9!
=> x^2 = 24 * 4! 7! / 2! 9!
=> x^2 = 24 * 4*3 / 9*8
=> x^2 = 4
=> x = 2 (assuming no calculation mistakes)
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suppose X = 2^n + 3^n + 5^n
now for n = 1, X = 10 (not div by 4)
Now for any n > 1, 2^n is always div by 4.
3^n gives remainder 1 when n is even
(if n is even, 3^n = 3^(2*n/2) = 9^n/2 = (8 + 1)^n/2 do binomial expansion n see all terms are div by 4 except last one.)
3^n gives remainder 3 when n is odd
(if n is odd, 3^n = 3 * 3^(2*(n-1)/2) = 3 * 9^(n-1)/2 = 3 * (8 + 1)^(n-1)/2 do binomial expansion n see all terms are div by 4 except last one i.e. 3.)
5^n gives remainder 1 for all n.
(5^n = (4 + 1)^n do binomial expansion n see all terms are div by 4 except last one i.e. 1.)
now for 2^n + 3^n + 5^n to be div by 4, n must be >1 and an odd number because then remainder of 3^n that is 3 and remainder of 5^n that is 1 will add and make 4. So whole number 2^n + 3^n + 5^n will be div by 4.
=> no. of ways of selecting = 50 - 1 = 49 Ans
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Vandana!
let z is a complex number as z = cos A + i sin A
=> z^n = (cos A + i sin A)^n = cos nA + i sin nA (De-Moivre's Theorem)
Now consider the series:
S = ( z * cos A) + ( z * cos A)^2 + ( z * cos A)^3 + ......+ ( z * cos A)^n ........ (i)
=> S = (cos A + i sin A)* cosA + (cos 2A + i sin 2A) * (cos A)^2 + ..................
.......................(cos nA + i sin nA)* (cos A)^n
Now imaginary part of S is,
Im (S) = sin A * (cos A) + sin 2A * (cos A)^2 + sin 3A * (cos A)^3 +.......................................sin nA (cos A)^n = given expression
but (i) is a GP and so S = [(z cos A) ((z cosA)^n - 1) ] / (z cos A - 1)
thus:
sin A * (cos A) + sin 2A * (cos A)^2 + sin 3A * (cos A)^3 +.................sin nA (cos A)^n = Im ([(z cos A) ((z cosA)^n - 1) ] / (z cos A - 1)) Ans
RHS can be found easily.
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Shashank,
you have to somehow distribute 2, 2, 5 and 3 among x1, x2, x3 as 2*2*3*5 = 60
now u can give all 2s to any one = 3 ways
or 1-1 2 to any 2 of them = 3 ways (to x1, x2 or to x2, x3 or to x3, x1)
=> no. of ways of distributing 2s = 6
Now 3 and 5 can be distributed either both to any one i.e. 3 ways or 1-1 to any two of them = 3 ways but they can be exchanged also i.e. 3*2 = 6 ways .... thus total 3 + 6 = 9 ways total to distribute 3, 5
=> total no. of ways = 6 * 9 = 54
is that clear?
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60 can be written as 2*2*3*5 = (2^2)(3^1)(5^1)
now x1, x2, x3 can be formed by taking two 2s, one 3 and one 5 => total 4 items
we have to effectively distribute these 4 items among three persons (x1, x2, x3) where we can 0 items to anybody too. (in that case x1, x2 or x3 is 1)
so no. of ways = 3 * 9 + 3 * 9 = 54 Ans
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Here u have to take 5 digits out of 6 digits, you can take either 0,1,2,4,5 or 1,2,3,4,5 only.
when digits are : 0,1,2,4,5
number of numbers = 4*4*3*2*1 = 96
when digits are : 0,1,2,4,5
number of numbers = 5*4*3*2*1 = 120
=> total no. of digits = 120 + 96 = 216 Ans
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number of 3 digit distinct nos whose sum of digits are even:
first digit can be anything from 1 to 9 => 9 ways
second can be anything from 0 to 9 => 10 ways
third can be 1,3,5,7,9 (if sum of first two digit is odd) or 0,2,4,6,8 (if sum of first two digit is even) => 5 ways
so total no. of ways = 9 * 10 * 5 = 450 Ans
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Akash!
S = 1.3.5.7 + 3.5.7.9 + 5.7.9.11 + ............. n terms
=  (2r-1)(2r+1)(2r+3)(2r+5) where r = 1 to n
i.e. S = Tr where Tr = (2r-1)(2r+1)(2r+3)(2r+5) and r = 1 to n
Tr = (2r-1)(2r+1)(2r+3)(2r+5)
or Tr = (1/10) [(2r-1)(2r+1)(2r+3)(2r+5)(2r+7) - (2r-3)(2r-1)(2r+1)(2r+3)(2r+5)]
=> T1 = (1/10) [ 1.3.5.7.9 - (-1).1.3.5.7]
T2 = (1/10) [ 3.5.7.9.11 - 1.3.5.7.9]
T3 = (1/10) [ 5.7.9.11.13 - 3.5.7.9.11]
T4 = (1/10) [ 7.9.11.13.15 - 5.7.9.11.13]
.......................................................
Tn = (1/10) [(2n-1)(2n+1)(2n+3)(2n+5)(2n+7) - (2n-3)(2n-1)(2n+1)(2n+3)(2n+5)]
(please note here that second part of every term is first part of previous term)
adding all of them,
S = T1 + T2 + T3 +......+ Tn
=>. S = (1/10) [(2n-1)(2n+1)(2n+3)(2n+5)(2n+7) - (-1).1.3.5.7]
=> S = (1/10) [(2n-1)(2n+1)(2n+3)(2n+5)(2n+7) + 105] Ans
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Please check the ques again and write it in a clearer way.
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Probability for the 1st item to be non-defective = (n-r)/n
Probability for the 2nd item to be non-defective (given that the 1st item was non-defective) = (n-r-1)/n
Probability for the 3rd item to be non-defective (given that the 1st n 2nd items were non-defective) = (n-r-2)/n
and so on.....
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If an item is put back in the lot after inspection,
probability = [ ((n-r)/n)^(k-1) ] * (r/n)
If an item is not put back in the lot after inspection (i.e. inspect and put aside),
For kth article inspected to be the first defective one, last k-1 items must come as non-defective,
prob = (n-r)/n * (n-r-1)/n * ............. (n-r-k+2)/n * r/(n-k+1)
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shashank!
you are right!
I apologize for the error.
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Pooja!
As far as I understood your question, i think you are asking the probability that 1, 2, and 3 come together in the order mentioned i.e. 1 then 2 and then 3.
Now, total number of arrangements are 9! of course.
and favorable no. of outcomes are when you consider the digits 1,2,3 as a single digit......then there are 6 digits + 1 digit (123) = 7 digits
so, number of favorable arrangements = 7!
so probability = 7!/9! = 1/72
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shashank!
Consider the following A.G.P.
S = x^r - 2x^(r-1) + 3x^(r-2) - 4x^(r-3) +.............+ (-1)^r * (r+1) * x^0
....... (i)
(Calculate S using sum of AGP)
and the following binomial expansion:
(1 + 1/x)^n = C0 + C1/x + C2/x^2 + ..........+ Cn/x^n
=> (1 + x^n)/x^n = C0 + C1/x + C2/x^2 + ..........+ Cn/x^n
..........(ii)
Multiplying RHS of (i) and (ii), we get
[x^r - 2x^(r-1) + 3x^(r-2) - 4x^(r-3) +.............+ (-1)^r * (r+1) * x^0]
* [C0 + C1/x + C2/x^2 + ..........+ Cn/x^n]
(Cr is ncr here)
If you compare the coefficient of x^0 in this expression, you get,
ncr - 2ncr-1+3ncr-2-4ncr-3+...............................(-1)r(r+1)
and the value of this will be coefficient of x^0 when you multiply S and
(1 + x^n)/x^n where S you know after doing sum of AGP.
For any clarifications, ask further.
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Sonia,
there is so much of drawing in solution of ur ques. Please write your email here or send me at ken.rohit@gmail.com.
I will send u the solution within 6 hrs.
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Pooja!
Of course if n > 365 , probability = 0
If n < 365, then
Probability that all n persons must have altogether different b'days is equal to
= (365/365) * (364/365) * (363/365) * .......................... ((365 - n + 1)/365)
= (365!) / ((365^n)(365 - n)!)
Now the probability that at least two of them have same birthdays
= 1 - [ (365!) / ((365^n)(365 - n)!) ]
Here, i have assumed that we are talking abt a non-leap year.
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if u solve x^3 = 1, u will get it, i m giving some steps as follows:
=> x^3 - 1 = 0
=> (x - 1)(x^2 + x + 1) = 0
=> x = 1, ( - 1   -3 ) /2
=> x = 1, ( - 1  i  3 ) /2 or 1, w, w^2
similarly if u solve given equation which is (x^2 - x + 1) = 0, u will get - w, - w^2.
please call or mail me for further queries. it's painful to type here.
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x + 1/x = 1
=> x^2 - x + 1 = 0
=> x = ( 1 3 i )/2 ( i is imaginary unit (iota) )
=> x = - w, - w^2 (w is omega, third root of unity, w^3n = 1 for any +ve integer n)
=> x^2000 = (- w)^2000 = ((- w)^2) (- w)^1998 = w^2 . 1 = w^2
=> x2000 +1/x2000 = w^2 + 1/w^2 = (w^4 + 1)/w^2 = (w + 1)/w^2
= - w^2/w^2
=> x2000 +1/x2000 = - 1 Ans
Please note that:
Since x + 1/x = 1, x can't be real, as for real x,
(x + 1/x)/2 >= (x. 1/x)^(1/2) = 1 (using A.M. >= G.M.)
=> x + 1/x >= 2 for real x.
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Sneha,
sin-1x can be max pi/2 and minimum -pi/2 (as far as i remember the range of sin-1x)
let A = sin-1(sin 5)
=> sin A = sin 5 = sin (pi + (5 - pi)) = - sin (5 - pi)
=> sin A = - sin (pi/2 + (5 - 3*pi/2)) = - cos (5 - 3*pi/2)
=> - sin A = cos (5 - 3*pi/2)
=> cos (pi/2 + A) = cos (5 - 3*pi/2)
=> pi/2 + A = 5 - 3*pi/2
=> A = (5 - 2*pi) is the correct answer.
(remember that answer must lie between -pi/2 and pi/2, and (5 - 2*pi) does)
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Vikram!
Best book is TMH Book but it's level is a little bit high for the beginners. Other good books are by A Das Gupta and Bharti Publication Multiple choice book.
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Vriti Edustore
