Home » Ask & Discuss » Mathematics. » Algebra « Back to Discussion

Algebra

Hot goIITian

Joined:	25 Jan 2009
Post:	119

3 Feb 2009 23:50:31 IST

0 People liked this

386

n balls are picked, one by one from four urns each containing n identical balls of green, red, white

None

n balls are picked, one by one from four urns each containing n identical balls of green, red, white and blue colours. The number of ways, when no two consecutive pickings are from same urn and balls of all the colours are picked,

Share this article on:

Comments (7)

~goiit user~

Hot goIITian

Joined: 25 Jan 2009

Posts: 119

4 Feb 2009 21:48:07 IST

Like 1 people liked this

replies pleaseeee

~goiit user~

Hot goIITian

Joined: 25 Jan 2009

Posts: 119

6 Feb 2009 21:43:36 IST

Like 1 people liked this

ne one trying ?????

Aatish Pandit

Blazing goIITian

Joined: 3 Feb 2007

Posts: 1200

6 Feb 2009 23:35:15 IST

Like 1 people liked this

Hey buddy..........first of all add a condition to ur question.....n >= 4....then now let us assume any n for simplicity to understand.....say n= 5.....then now for selecting 1st ball......4 options......for 2nd.....3 options.........for 3rd again 3optns.....for 4th again 3 optns.....for 5th again 3 optns.....so no of ways........4C1 x 3C1 x 3C1x 3C1x 3C1so for general n........nC1 x (n-1)C1 x (n-1)C1x (n-1)C1x (n-1)C1......where (n-1)C1 is multiplies n-1 times,.........i hope its clear........

RyuAmakusa

Blazing goIITian

Joined: 21 Mar 2008

Posts: 776

6 Feb 2009 23:54:35 IST

Like 0 people liked this

this indeed does takes care of pcking n balls no cons. from the same urn......but does this include selecting balls of all colour since balls of all the colours need to be selected and say the first four are of 4 diff colours. then for the 1st ball u have 4 urns to select andonce the urn is fixed u have 4 choice of colours so 4*4 for the second ball 3*3 3rd 3*2 4th 3 now for the remaining n-4 balls 3 options each......so 4!*4*3^n-1........this is true if the order of selection does not matter.........

~goiit user~

Hot goIITian

Joined: 25 Jan 2009

Posts: 119

16 Feb 2009 15:12:11 IST

Like 1 people liked this

Aatish your answer is wrong

please help !!!!!!!!!

~goiit user~

Hot goIITian

Joined: 25 Jan 2009

Posts: 119

17 Feb 2009 08:18:00 IST

Like 1 people liked this

some one help please

Kevin Arnold

Cool goIITian

Joined: 15 Jan 2009

Posts: 73

17 Feb 2009 15:53:35 IST

Like 2 people liked this

the number of ways of picking up the first ball is 4, the number of ways of picking up the second one is 3 (leaving the color picked first), the third one can be picked up in another 3 ways (leaving the color picked up in the 2nd pick) and so on ....we have

4.3^n-1......1) ways of picking up such balls...but we have not ensured that balls of all colours are picked up....only ensuring that alternating colored balls are picked up.

total number of ways in which we can make the same procedure with just 3 colors is agin done the same way...

there are 4C3 ie 4 ways of choosing those 3 colors from 4 colors...first ball in 3 ways second in 2 ways and so on...we have 3.2^n-1.....

so 4.3.2^n-1.........2)

now the total number of ways in which we can make the same construction with 2 balls is...

there are 4C2=6 ways of choosing 2 colors from 4 colors...

and the same ways the first ball can be picked in 2 ways and the rest of the n-1 balls in 1 way each...

so 6.2 ways........3)

now by inclusion exclusion principle we have the total number of such selections is

1)-2)+3)

=4.3^n-1-12.2^n-1+12

=12(3^n-2-2^n-1+1) ways

cheers!!!

Quick Reply

Reply

	Some HTML allowed.
	Keep your comments above the belt or risk having them deleted.
	Signup for a avatar to have your pictures show up by your comment
	If Members see a thread that violates the Posting Rules, bring it to the attention of the Moderator Team