24Q7 x=(2n+1)(2n+3)...(4n-3)(4n-1);y=(\frac{1}{2})^n\frac{(4n)!.n!}{(2n)!(2n)!}
find value of x-y+2n
Q1 no of six digit nos which have sum of digits as odd integer
Q2 find Ak if \frac{m!}{x(x+1)(x+2)..(x+m)}=\sum_{i=1}^{m}{\frac{A_i}{x+i}}
Q3 largest two diigt prime number that divides^{200}C_{100}
Q4 no. of natural nos n for which n! ends with 26 zeroes
Q5 find no. f 3 digit nos .of the form xyz where x,z<y ,x≠0
Q6 what are ternery sequences ??
how many terenry sequences of length 9 are there which begin with 210 or end with 210
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6 Answers
24
106Q4. we need to find where the exponent of 5 = 26
till 100 it is 20+4 = 24..
now 105 has 25
and 110 has 26
so 110!, 111!, 112!, 113!, 114! have 26 zeroes
21Q3
200C100=200!100!.100!
in 200! der r exactly 2 nos which r multiple of 67,68,69.......100
whearas in 100! der is one multiple of 67,68......100...i.e the no. itself
des nos 67,68,69......100 gets cancellled from numerator & denominator in val of 200C100
in 200! der r exactly 3 nos which r multiply of 51,52..........66
wheras in 100! der is only one number multiple of 51,52....66(i.e no. itself)
so we hav 200C100=(61)3(No. not divisble by 61)(61x No. not divisble by 61)2
=61x integer
hence 61
1What is the sum of the digits
a > in 100000 --- ans. 1
b> in 100001 --- ans. 2
Take any arbitrary ,
c > in 123243 --- ans . 15
d > in 123244 --- ans . 16
So numbers which have odd sum of digits come right behind those who have even sum of digits .
What I want to say is , that the number of six digit numbers , that have odd sum of digits , is half the total no. of six digit nimbers.
So ans . --- 500000
1Given x = (2n+1)(2n+3) ........ (4n-3)(4n-1)
So x . (2n+2) . (2n+4) ..... 4n = (4n)! / (2n)!
Here we have n terms in L . H . S, starting from 2n+2 to ending with 4n . So by taking 2 from
each term , we get 2n [ (n+1)(n+2)(n+3) …. Upto 2n ] , which is (2n)! . 2n / n!
So x . (2n)! . 2n / n! = (4n)! / (2n)!
or x = 2-n . n! . 4nC2n -------------------- 1
Again , y = 4nC2n . n! . 2-n ------------- 2
So from 1 and 2 ,
y = x
Finally , x - y + 2n = 2n