IIT JEE , AIEEE Forum - Mathematics , Algebra

rishipratimm

HEY GUYS...

TEST UR SKILLS WID THESE QUESTIONS....

RATES FOR COMPLETE SOLNS!!!!!

Q1. THERE ARE 3 NON ZERO POSITIVE INTEGERS a,b,c
        GIVEN : LCM OF a,b   = (2^3)(5^3)
                         LCM OF a,c   = (2^4)(5^3)
                         LCM OF b,c    =(2^4)(5^3)
        HENCE FIND THE TOTAL NUMBER OF TRIPLETS (a,b,c)

Q2. A SUBSET A IS CHOSEN FROM A SET CONTAINING n ELEMENTS
       THEN THE ELEMENTS OF SET A ARE REPLACED INTO THE ORIGINAL SET.
        FIND THE NUMBER OF CASES WHEN
         i)AUB CONTAINS r ELEMENTS
        ii)AUB IS THE UNIVERSAL SET
        iii)n(B)<n(A)

Q3.THE LETTERS OF THE WORD "SUCCESS" ARE REARRANGED.
FIND THE TOTAL NO. OF WORDS FORMED SUCH THAT NO TWO CONSEQUETIVE LETTERS ARE NEXT TO EACH OTHER.

Q4.A LINE OF LENGTH a IS DIVIDED INTO 3 PARTS.
FIND THE PROBABILITY THAT THE THREE PARTS FORM THE SIDES OF A TRIANGLE.
i) 2/7 ii) 3/13 iii) 1/4 iv) 3/7

Q5.THE SUM OF THE COEFFICIENTS OF INTEGRAL POWERS OF x IN THE EXPANSION OF [1+2(x^1/2)]^40 is
_______

Q6.IN THE EXPANSION OF (x-1)(x^2 - 2)(x^3 - 3)...........(x^n - n)
FIND THE COEFFICIENT OF x^[(1/2)(n^2 + n + 14)]

akhil_o

to Q 4)

for triangle

sum of two sides>sum of third side

draw line a

dividin in 3 parts is same as choosing 2 points

now for x+y>z

the length of one side is always less than a/2

hence we choose point x from [0,a/2)

now from x we choose second point from (x,a/2+x] so that distance between last points also <a/2

so probability

=(a/2)(a/2)/a*a

=1/4

akhil_o

3)assuming u meant no 2 same letters shd be consecutiv...

we'll find the cases we dont want

we dont want cases in which any 2 SS or CC are together...note that SSS appearing together comes under SS case..and we also subtract cases with SS and CC togetther

SS together-(SS)CCUES

=6!/2

CC tog- (CC)SSSUE

=6!/3!

CC and SS tog- (CC)(SS)SUE

=5!/2*2

so no of unwanted cases=(360+120-30)=(450)

Subtract this from total no of cases

VIV

edited.

rishipratimm

dont u hink that when u r considerin the cases when only s are together,the case when c are also together also gets included?
similarly the case where c are together,there is a case where all the s are together,and several cases when two of the s are together...

rishipratimm

@akhil...
dont u hink that when u r considerin the cases when only s are together,the case when c are also together also gets included?
similarly the case where c are together,there is a case where all the s are together,and several cases when two of the s are together...

akhil_o

No...see...i hav subtracted cases when both ss and cc are togteher..
and when we choose two SS together.then we hav already counted cases of SSS so i didnt count that case at all
maybe there is a calculation error poss..plz check that

nadeemoidu

****edited********

elastiboysai

Hi guyz

I think this method is better.

Let us divide line AB of length a into 3 parts by 2 points say I and J.

let AI=x,IJ=y den JB=a-(x+y)

x+y<a

x+y>a-(x+y)

x+y>a/2

x+(a-x-y)>y

y<a/2

similarly x<a/2

Now just sketch these inequalities on a graph

Total possible cases are represented by the area of the big triangle formed by x+y<a and the axes

and favorable cases by

the area of the triangle formed by x+y>a/2 x<a/2 and y<a/2

so Probability of formin a triangle =

[a^2/8]/[a^2/2]

gives 1/4

akhil_o

not sure of this...

1)

analysing data,we can see that a b and c are multiples of 5 and 2 only

also since LCM of a and b has 2 power 3, none of a and b can have term 2^4

since ac and bc LCMs have 2^4 term, c MUST have 2^4 term

so for shuffling powers of 2 between a and b

we have a=3,b=0,1,2,3 or vice versa

hence 2*4 sets=8 possiblities

now for power 5...

we can see that to maintain an LCM of 5^3 for all three cases,

we need to keep atleast 2 terms with power 5^3

other can take powers 0,1 or 2

so for shuffling pors of 5 we have 3*3=9 sets

so number of possibilities is

8*9=72

akhil_o

@nadeemoidu, i ddnt mean that...we r not dividin the line in two parts...
after we fix point x, the next a/2 i am talking about is the segment of length a/2 from x as starting point...so a triangle will always be possible..will post a diagram to make iit clear

rishipratimm

guys .....
i solved the first sum in this way....

let us suppose that

a = (2^x)(5^y)

b= (2^q)(5^r)

c=(2^p)(5^s)

rishipratimm

now from the given conditions,we get
max(x,q)=3
max(p,x)=4
max(p,q)=4
thus using this condition we get....
x q p
3 0 4
3 1 4
3 2 4
0 3 4
1 3 4
2 3 4
3 3 4
in this way,we get the total no. of combinations of the powers of 2 ....
using a similar method to calculate the combinations of powers of 5,and multiplying the two cases,we can get the total no. of cases....

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