IIT JEE , AIEEE Forum - Mathematics , Algebra

rishipratimm

the number of garlands which can be made by 7 flowers of one kind and 2 flowers of anothr kind must be

i)2^7

ii)9!/7!2!

iii)4

iv)6

Ronaldooo

Its 4...its given in fiitjee key that its number of ways of 3-partition of 7..even i didn understand the logic...

rishipratimm

well...
i too am confused by dat....
apart from the partitions givn in fiitjee,i think 2 more partitions shud also b there....
sum1 pls ans this q!!

ashish_banga

the answer acc to me should be 2 (not second option)

srini

Hey,

ya it should be(9-1)!/7!2!*2

rishipratimm

@srini & ashish
pls cud u giv d reasons for ur ans???

ashish_banga

exactly the same reason as of srini
the answer is 2

srini

Hey,

in a circular permutation,

We have in linear 1 2 3 ...n

2 3 4...n 1

3 4..n 1 2

.

n times

goes on... so in circular we consider all these equal hence the answer of (n-1)! comes

Now we have certain equal quantities which we have taken that many factorial times so we divide by the same

.

Ronaldooo

I think you'll also have to multiply it by 2 as it can also be anticlockwise...thats y maybe answer is given as 4..

RyuAmakusa

ans is 4. its very easy.
first arrange the 7 flowers.there is only 1 way.
now when you r putting the other 2 there can be 0,1,2,3 flowers
of the 7 kind . which accounts for 4 cases

rishipratimm

THANKS,GUYS!!!!

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