
The value of the expression 47C4 + j=15Σ52j C3 IS ... ...

Find the H.C.F. of the minimum nonnegative values of a,b and c, given that the equation x4+ax3+bx2+cx+1=0 has only real; roots ... ...

For positive integers n1 and n2 , the value of the expression (1+i)n1+(i+13)n1+(i+15)n2+(i+17)n2 here i= 1 is real if and only if (A)n1=n2+1 (B) n1=n21 (C) n1=n2 (D)n1>0,n2 >0 pl post the soln as well . ...

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*Image* a)For the first part the expected value would be = E(x)=\frac{1}{2}+2\left(\frac{1}{2}\right)^2+3\left(\frac{1}{2}\right)^3+\cdots=2 b)For the second part the expected value would be = E(x)=\frac{1}{2}+\frac{2}{4}+\fr ...

The solution set of frac{{{x^2}  3x + 4}}{{x + 1}} > 1,,x in R , is a) (3,,, + infty ) b) (  1,,,1) cup (3,,, + infty ) c) [  1,,,1] cup [3,,, + infty ) d) None of these ...

3.nc08.nC1+13.nC218.nC3.......... upto (n+1) terms =? ...

Find the coefficient of xr in the following (x+2)n+(x+2)n1(x+1)+(x+2)n2(x+2)2+.....+(x+1)n ...

The probability that a randomly chosen relation from a set A = {1, 2, . . . n}, n ≥ 1 to itself is a symmetric relation is (A) 2 n2+n/2 (B) 2 n2/2 (C) 2 nn2/2 (D) 2 n+1n2/2 ...

The value of i log(x – i) + i2+i3 log(x +i) + i4( 2 tan1x), x> 0 is (A) 0 (B) 1 (C) 2 (D) 3 ...

The product of n positive numbers is unity.Then their sum is a.a positive integer b.divisible by n c.equal to n + 1/n d.never less than n Let the numbers be a1,a2,a3.......an Applying A.M>=G.M we get, a1+a2+a3+............ ...

Consider a 25*25 grid of city streets. Let S be the points of intersection of the streets, and let P be the set of paths from the bottom left corner to the top right corner of which consist of only walking to the right and up ...

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in how many ways we can make a 4 digit no. using 1,2,3,4 and what's the sum of all the numbers obtained?(repeatations not allowed) ...

If a,b,c be the pth,qth and rth terms of an HP respectively then prove the system of equations (bc)x + (ca)y + (ab)z = 0 px +qy + rz =0 and x+y+z=0 has ∞ number of solutions ...

If a,b,c are in AP ;b,c,a In GP then prove that 1/c , 1/a , 1/b are in ap ...

This is a great problem which you can probably approach in a number of different ways. All natural numbers are arranged as follows. 1 3 6 10 15 . 2 5 9 14 . . 4 8 13 . . . 7 12 . . . . 11 . . . . . and so on. Find the row and ...

Three distinct numbers are selected uniformly at random from the tenterm geometric sequence with first term 10/9 and common ratio 2 . What is the expected value of their sum? ...

Find till infinity: 1  n^2 + {n^2(n^21^2)}/(2!)^2  {n^2(n^21^2)(n^22^2)}/(3!)^2................................ to infinity,where n belongs to N. ...

If each pair of the following three equations x2+ax+b=0, x2+cx+d=0, x2+ex+f=0 has exactly one root in common,then show that (a+c+e)2=4(ac+ce+eabdf). ...

A box contains coupons labeled 1,2,3,....n. A coupon is picked at random and the number x is noted. The coupon is put back into the box and a new coupon is picked at random. The new number is y. Then the probability that one ...

Let S={1,2,3,...,n} and A={(a,b)l1≤a,b≤n}=S X S. A subset B of A is said to be a good subset if (x,x) belongs to B for every x belonging to S. Then the number of good subsets of A is A. 1 B. 2n C.2n(n1) D.2n2 The answer ...

\hspace{16}$Find all real polynomials $\bf{p(x)}$ such that $\bf{p(x)\cdot p(x+1)=p(x^2)\;\forall x\in \mathbb{Z}}$ ...

The maximum value M of 3x+5x9x+15x25x,as x varies over reals,satisfies A.3<M<5 B.9<M<25 C.0<M<2 D.5<M<9 ...

\hspace{16}$In how many ways can the selection of $\bf{8}$ letters be done frm $\bf{24}$ letters\\\\ of which $\bf{8}$ are $\bf{'a'}$ and $\bf{8}$ are $\bf{'b'}$ and rest are unlike. ...

Three children,each accompanied by a guardian,seek admission in a school.The principal wants to interview all the 6 persons one after the other,subject to the condition that no child is interviewed before its guardian.In how ...

If 'a' is a complex number such that a=1.Find the values of a,so that equation az2+z+1=0 has one purely imaginary root. ...

S=11+11+11....................... infinity now thus, S=1 (11+11+11....................... infinity) thus S=1S thus 2S=1 S=1/2 ...

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There are 5 different red balls,5 different green balls,5 different blue balls and 5 different black balls.In how many ways can they be arranged so that no two balls of same color are adjacent ? ...