IIT JEE , AIEEE Forum - Mathematics , Algebra

santosh05

how many no xyz can b formed witd x<y and z<y

akhil_o

Choose any three numbers from [1,9] without repetition
each such combination forms a number if arranged in asc order
so answer=9C3

=84

Conjurer

Middle term is largest akhil-o

A long way will be to make cases and see.

akshay.khare91

here it means that y is greatest..

so the cases can be formed by fixing the value of y and then observing in how many ways the x and z can be selected..
if 2 is fixed at y place then x can be filled by 1 and z can be filled by 0 or 1 ...
x y z
1 2 2 = 2

similarly fixing different nos we see that

2 2 3 = 6
3 3 4 =12
4 4 5 = 20
5 5 6 = 30
6 6 7 = 42
7 7 8 = 56
8 8 9 = 72

adding all these ways we get = 240 -- ans...

computer001

y=2 : 1*2
y=3: 2*3
y=4:3*4
so it will be 1*2 + 2*3 + 3*4 + 4*5 +5*6 +6*7 + 7*8 +8*9=
2+6+12+ 20 + 30 + 42 + 56 + 72= 240

a5hw1n_5

take different cases
if y=9 then x is bet 1-8 and z bet 0-8
therefore no of ways of choosing x and z is 9(9-1)
for y=r
no of ways is r(r-1)
hence tot no of ways is= _[1]

^{[ 9]} [r(r-1)]
=_[1]

^{[ 9]} [r²-r]=n(n+1)(2n+1)/6 - n(n+1)/2
=285-45=240

santosh05

hey ashin....can u pl xpaln d thrd n fifth line pl???

a5hw1n_5

if y=9 x can take value 1,2...or 8 ie there are 8 ways of choosing x. similarly z can take values 0,1,...or 8 so there are 9 ways of choosing z
hence the total number of ways of choosing x and z if y=9 is 9*8= 9(9-1)
therefore if y=r (1<r<=9) then the total number of ways of choosing x and z is obviously r(r-1)

so as y takes the values from 2-9 the number of combinations is given as
_[2]

^{[ 9]} [r(r-1)]
i given the previous ans as _[1]

^{[ 9]} [r(r-1)]. but u get the same ans anyways as 1(1-1)=0.
note- y cant take the val 1 as x cannot be 0 and x<y.
hope i cleared it for u. cheers!!

santosh05

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