#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-5470-145

+91 7353221155

CART 0

• 0
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

Ojas Jain
14 Points
9 years ago
Look either you can go by:-

Select any bead first- no of ways:6C1

Then arrange the others now because the symmetry is broken-no of ways:5!

But after writing this u will find that the same structure would be repeated because of the circular arrangement when u take another beads as starting point as thus forms a circular arrangement and thus it never end. It is a continuous arrangement
Thus every structure is repeated 6 times, then divide throughout by 6

The Ans is: 6x5!/5= 5!

OR
Another method for all such garland sums are (n-1)! Ehich is always valid for any of the circular arrangement
akash aggarwal
19 Points
9 years ago

5/2!

Anubhav Sharma
5 Points
9 years ago

accordingto you answer is coming 120 , but the actual answer given in my book is 60

Anubhav Sharma
5 Points
9 years ago

Ojas Jain
14 Points
9 years ago

That is becoz after getting 5! We see that the clockwise and anticlockwise arrangement is same

Thus the ans is 5!/2