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..!..THEORY ANSWERS..!..

(1)
(Log(base 2)m)^2-log(base
2)m^5=10
(log(base 2)m)(log(base
tita)m)-3log(base 2)m=10
(log(base 2)m)(log(base 2)m-3)=10
(log(base 2)m)9log(base 2)m-3)=10
log(base 2)m-3=10
log(base 2)m-3=10/log(base 2)m
log(base 2)m-log(base
2)m^3=10log(base 2)m^-1 log(base 2)m-3log(base 2)m=-10log
(base 2)m
log(base 2)m/(3^-1)=log(base 2)m^-10
m/3^-1=-10
m=-10*3^-1
m=-10*1/3
m=-10/3

Source: SchoolMaster dot teeKay
(2a)
given m*n =((m^2)-(n^2))/2mn
-3*2=(-3)^2-(2)^2
=(9-4)/-12
=5/-12
=-5/2

(2b)
to show that whether * is
associative
=(m*n)*p=m*(n p)
=(m n)*p=((m^2)-((n^2))/2mn)*p
=((m^2)-((n^2)^2-p^2)/(((m^2)-
((n^2))/2mn)p
=(((m^2)-((n^2)^2)/4(m^2)(n^2))-
p^2)/2p((m^2)-(n^2))/2mn)
=(m^2)-((n^2)^2)/4(m^2)(n^2)
(p^2)/4(m^2)(n^2))* 2mn/(2p
(m^2)-(n^2))
=((m^2)-(n^2))^2-(2mnp)^2)/(2mn
(2p)((m^2)-(n^2))
=((m^2)-(n^2))^2-(2mnp)^2))/4pmn
(m^2)-(n^2))
also, m*(n p)=m*((n^2)-(p^2))/2np)
=m^2-((n^2)/2np)^2)/2m((n^2)-
(p^2))^2/2m((n^2)-(p^2)/2np))
=(m^2)-((n^2)-(p^2))^2)/4np))/(m)
((n^2)-(p^2))/np
=(4(m^20np-((n^2)-(p^2))^2)/4np)/
m((n^2)-(p^2))/np
=(4(m^20np-((n^2)-(p^2))^2)/4np)/
m((n^2)-(p^2))/np
=(4(m^20np-((n^2)-(p^2))^2)/4np)*
(np)/m((n^2)-(p^2))
hence * is not associative

(6)
No of man =5,
No of woman =3,no of committee
=3 total no of people =5 3=8.
A no of ways of foring the committee=8C base3,
but nC base r=n!/(n-r)!r!,
8C base3=8!/(8-3)!3!,
=8!/5!3!= 8*7*6*5/5!*3*2=56ways.

6b.
prob ( of at least one woman is on
the committee)
= 5C2 3C1/8C5 8C3
= 5!/2!(5 - 2)! 3!/11(3 -1)!= 5!/2!3!
3!/1!2! = 120/2X6 6/1X2
8!/5!(8-5)! 8! /3!(8-5)! = 8!/5!3!
8!/3!3! = 40320/120X6 40320/6X6
= 120/12 6/2=10 3 = 13
40320/720 - 40320/36 = 56 1120 =
1176

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