Thank you for registering.

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

Please check your email for login details.
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

There are 3 piles of books on the table you need to arrange them in the self such that 1st book of each pile gets placed on the self,what is the possible no of arrangements: a)12! b)12C3 * 9C4 c)12^12 d)none

There are 3 piles of books on the table you need to arrange them in the self such that 1st book of each pile gets placed on the self,what is the possible no of arrangements:
a)12!    b)12C3 * 9C4 c)12^12 d)none

Grade:

2 Answers

basit ali
41 Points
10 years ago

3 piles i.e.  we have to select 1 book from each piles,

 3C1*3C1*3C1

=( 3!/1!*2! ) 3

 =3*3

=9   and hence answer is (D)

sharad akhouri
14 Points
9 years ago

There can be 4 powers of 7 which will give the last 4 digits as :

7^1 = 7

7^2 = 9

7^3 = 3

7^4 = 1

 

Other powers will be of the form: 4k, 4k+1, 4k+2, 4k+3. And all the four have an equal probability of being selected.

So, total number of combinations = 4x4 (repetition being allowed)

Total number of favourable combinations = (7,3)(3,7)(9,1)(1,9) - 4

So probability = 4/16 = 1/4

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free