IIT JEE , AIEEE Forum - Mathematics , Algebra

kishan12

A shopkeeper sells three variety of shampoo and he has a large number of bottle of the same size of each variety in his shop.there are 5 places in a row in his showcase. the no of different ways of displaing the three varity of shampoo in the showcase
Ans is 150 but i am getting 120...........................

Plz solve this without making cases.

akhil_o

1-1-3
3C1*5!/3!
=3*4*5=60

1-2-2
3C1*5!/2!2!
=3*120/4=90

so 150

1-1-3 means 3 of 1 type, 1-1 of others

kishan12

what i did is
x + y+ z=5
1<=x<=3
1<=y<=3
1<=z<=3

now we want coficient of x^5 in (x+x62+x^3)^3

i am getting 120................plz help me.........i wasted a hell of time on this

akhil_o

this just gives no of waus of choosing i think...not arranging

kishan12

arey then we permute it na.................

akhil_o

but to permute it u need exactly the case:
for eg if u have AAABC u cant permute it as 5!
it changes for each case!

kishan12

right but we can do it this way that
6 * 5!/3! as we know that 3 types of same bottles are going to caome.
(6 is no of ways of combi.)

konichiwa2x

no what u said in your previous post is wrong. Dividing by 3! is wrong. if the arrangement is : 1 2 2 then divide by 2!.2! as there are 2 objects of each type of shampoo and if it is 1 1 3 then divide by 3!

kishan12

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