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 Discussion Forums -> This Post 5 points    (1    in 1 votes )   [?]
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)
 Discussion Forums -> This Post 5 points    (1    in 1 votes )   [?]
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

 Discussion Forums -> This Post 27 points    (5    in 6 votes )   [?]
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.

 Discussion Forums -> This Post 20 points    (4    in 4 votes )   [?]
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?
 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]
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

 Discussion Forums -> This Post 5 points    (1    in 1 votes )   [?]
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

 Discussion Forums -> This Post 12 points    (2    in 3 votes )   [?]
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
 Discussion Forums -> This Post 39 points    (7    in 9 votes )   [?]
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
 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]
Please check the ques again and write it in a clearer way.
 Discussion Forums -> This Post 5 points    (1    in 1 votes )   [?]
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.....

 Discussion Forums -> This Post 5 points    (1    in 1 votes )   [?]
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)

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]
shashank!

you are right!

I apologize for the error.
 Discussion Forums -> This Post 5 points    (1    in 1 votes )   [?]
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

 Discussion Forums -> This Post 15 points    (3    in 3 votes )   [?]
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  nc 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.

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]
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.
 Discussion Forums -> This Post 2 points    (0    in 1 votes )   [?]
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.
 Discussion Forums -> This Post 5 points    (1    in 1 votes )   [?]
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.
 Discussion Forums -> This Post 7 points    (1    in 2 votes )   [?]
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.
 Discussion Forums -> This Post 10 points    (2    in 2 votes )   [?]
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)
.
 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]
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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