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Permutation combination
by Saurabh Kumar - Saturday, 6 September 2008, 07:50 PM
 

Q1:How many 6 digit numbers exactly contain 4 different digits?

 

Q2.How many numbers smaller than 2*10^8 and divisible by 3 can be written by means of 1,2 and 0(excluding single and double digit nos.)?

 

 

 

Q3.SIX WHITE AND SIX BLACK BALLS OF SAME SIZE ARE DISTRIBUTED AMONG 10 URNS SO THAT THERE IS AT LEAST ONE BALL IN EACH URN.WHAT IS NUMBER OF DIFFERENT DISTRIBUTIONS OF BALL? 

Re: Permutation combination
by vamsi krishna - Saturday, 6 September 2008, 08:43 PM
  Saurabh

here goes my answers....

1) 9C4.4^6 + 9C3.3.4^5
2) I cudnot do that in 5 mins
3) 25200
Re: Permutation combination
by Saurabh Kumar - Saturday, 6 September 2008, 09:46 PM
 

pLZ GIVE DETAILED EXPLAINATIONS....THANKS..I cud not understand..how u got it

Re: Permutation combination
by harry singh - Sunday, 7 September 2008, 01:20 AM
 

First, an observation- In base with b, there are bn – 2 n-digit numbers beginning with any particular digit (not equal to 0) and divisible by b. For example in base 5, there would be 54 6-digit numbers beginning with 1, 54 6-digit numbers beginning with 2, 54 6-digit numbers beginning with 3,… 54 6-digit numbers beginning with 4 which are divisible by 5. Let’s see the proof of this observation:

Observation 1: One out of every n consecutive numbers is divisible by n.

Observation 2: Consider a base b. If you need to form n-digit numbers in base b, the first place can be filled in b - 1 ways (you cannot use 0) and the remaining n - 1 places can be filled in b ways. Remember that in base b, you have b digits, 0 to b - 1. Therefore, the total n-digit numbers you can form = (b - 1) Ã— bn - 1 . Since these would be consecutive numbers, the number of numbers divisible by b is = (b - 1) Ã— bn - 1 /b = (b - 1) Ã— bn - 2 . Out of these (b - 1) Ã— bn - 2 numbers, bn - 2 will be beginning with 1, bn - 2 will be beginning with 2, …, bn - 2 will be beginning with b - 1.

When you write numbers using only 0, 1 and 2, you are writing in base 3.

Therefore, 3- digit numbers in base 3 beginning with 1 and divisible by 3 = 3

4 – digit numbers in base 3 beginning with 1 and divisible by 3 = 9

5 – digit numbers in base 3 beginning with 1 and divisible by 3 = 27

6 – digit numbers in base 3 beginning with 1 and divisible by 3 = 81

7 – digit numbers in base 3 beginning with 1 and divisible by 3 = 243

8 – digit numbers in base 3 beginning with 1 and divisible by 3 = 729

9 – digit numbers in base 3 beginning with 1 and divisible by 3 = 2187

There will not be a 9- digit number beginning with 2 and less than 200000000.

Therefore, 3- digit numbers in base 3 beginning with 1 and divisible by 3 = 3

4 – digit numbers in base 3 beginning with 2 and divisible by 3 = 9

5 – digit numbers in base 3 beginning with 2 and divisible by 3 = 27

6 – digit numbers in base 3 beginning with 2 and divisible by 3 = 81

7 – digit numbers in base 3 beginning with 2 and divisible by 3 = 243

8 – digit numbers in base 3 beginning with 2 and divisible by 3 = 729.

Therefore, total number of numbers = 2(3 + 9 + 27 + 81 + 243 + 729) + 2187 = 4373.

Re: Permutation combination
by Total Gadha - Sunday, 7 September 2008, 11:12 AM
  smile
Re: Permutation combination
by vamsi krishna - Sunday, 7 September 2008, 01:52 PM
  U r goin great guns harry

U hav enlightened us !

ThankU
Re: Permutation combination
by vamsi krishna - Sunday, 7 September 2008, 01:57 PM
  @saurabh
Pl cnfrm Ans
Re: Permutation combination
by Saurabh Kumar - Sunday, 7 September 2008, 04:26 PM
  answer correct..but can someone help with question 1 and 3,,,,,,,,with detailed explaination
Re: Permutation combination
by vamsi krishna - Sunday, 7 September 2008, 04:42 PM
  Saurabh,

3) select five urns out of 10 urns to put 5 white balls = 10C5
put five black balls in remaining 5 urns in 1 way
u r left with one 1 white and 1 black ball
place white ball in any of urns in 10 ways
place black ball in any of urns in 10 ways

hence 10C5 . 10 . 10 ans

1)

divide into 2 cases.
a) '0' is not in da four digits
_ _ _ _ _ _
select 4 digits out of [1,9] in 9C4 ways and then each of space can be filled with one of these four digits in 4^6 ways
so ways = 9C4.4^6

b) '0' is included in da four digits
_ _ _ _ _ _
0 is already selected
select remaining 3 out of [1,9] in 9C3 ways
First space can be filled by one of the three selected digits in 3 ways ( it cant take zero)
then remaining 5 spaces can be filled with one of these 3 digits or o in 4^5 ways
so ways = 9C3.3.4^5

sum each case to get ans.

VaMsI
Re: Permutation combination
by Top CAT - Monday, 8 September 2008, 06:34 AM
  Hi Vamsi,

Q3. Your ans does not contain distribution containing 3 White or 3 black balls in any one urn
Re: Permutation combination
by vamsi krishna - Monday, 8 September 2008, 09:16 AM
  @Top Cat

Thanks Atom for pointing out...

How to approach then?

Sourav Pl give answers
( so that we can try n fit solution )

Re: Permutation combination
by sunil garg - Monday, 8 September 2008, 11:11 AM
 

divide into 2 cases.
a) '0' is not in da four digits
_ _ _ _ _ _
select 4 digits out of [1,9] in 9C4 ways and then each of space can be filled with one of these four digits in 4^6 ways
so ways = 9C4.4^6 

HI vamsi

i have doubt the way you have written " a six digit number with exactly four different digits can be written as 9c4x4^6......

the later part 4^6 .. i think is not okie.......say i have digits 1,2,2....according to ur formula the possible 2^3 =8 numbers will be there but in actual there are only 3 such numbers (122,221,212..)

isn't it

please correct me if i m wrong

regards

sunil garg

Re: Permutation combination
by vamsi krishna - Monday, 8 September 2008, 01:38 PM
  @ Sunil

i was forming six digit number using four digits..

take simple case

u have two non-zero digits say 1,2 and u need to form a 3-digit number(repetitions allowed)
possible cases are
111
112
121
122
211
212
221
222---8 cases hence 2^3


(why are you specifying those three digits..as 1,2,2,and taking permutations of that?
u jst need to form 3 digit number out of 1 and 2)

similiarly you have four nonzero digits and you need to form a 6 digit number - 4^6 possibilities

VaMsI
Re: Permutation combination
by Saurabh Kumar - Tuesday, 9 September 2008, 12:37 AM
  answer is wrong for question 1.correct answer is 191520..

answer is wrong for question 3.correct answer is 26250....
Re: Permutation combination
by Bishweshwar Pradhan - Tuesday, 9 September 2008, 11:54 AM
 

(3) . am getting different answer. here is my approach.

First divide 10 balls among 10 urns & keep aside 2 balls. These can be done in 3 ways. And the number of ways :-

1. 5 white & 5 red balls --> ( 10! /5! * 5! )

2. 6 white & 4 red balls --> ( 10! / 6! * 4! )

3. 6 red & 4 white balls --> ( 10! / 6! * 4! )

After dividing these 10 balls we can divide the rest 2 balls in 10c2 ways.

So total is  

[ ( 10! / 6! * 4! ) + ( 10! / 6! * 4! ) + ( 10! /5! * 5! ) ]  * 10c2 =  11340.

Correct me if any mistake.

Re: Permutation combination
by Himanshu Shekhar - Tuesday, 9 September 2008, 10:02 PM
 

Regardng problem 1, answer by vamsi krishna has considered many numbers which contain less than 4 digits, if 4^6 or 4^5 type statement is considered. There can be a number containing only single digit, 2 digits or 3 digits.

Although I am also quite away from given answer and my approach is slightly different and lengthy, but let me get feedback on my approach.

First case: No zero in the 6 digit number. Select 4 digits out of the remaining 9 digits (1 to 9) in 9C4 ways. At first, any two digits out of already selected 4 digits are taken for repetitive use in 4C2 ways. Now we have 6 numbers of the form aabbcd, which can be arranged in 6!/2!2! ways i.e 9C4 x 4C2 x 6!/2!2! = 136080. In this case, one digit out of already selected 4 can be selected for use three times in 4C1 ways to form numbers of the form aaabcd. total number of arrangements is 6!/3! i.e. 9C4 x 4C1 x 6!/3! = 60480.

Second case : One zero is selected. Select other 3 digits in 9C3 ways. At first select 2 out of selected 3 digits for repition twice (0aabbc) and reduce those conditions, when zero appears at leftmost poisiton. 9C3 x 3C2 x (6!/2!2! - 5!/2!/2!) = 37800. Another case like above, when one number is used thrice (0aaabc). 9C3 x 3C1 x (6!/3! - 5!/3!) = 25200.

Third case : Two zeros are selected. Selected 3 out of 9 digits in 9C3 ways. Selected 1 out of selected 3 digits for repition in 3C1 ways (00aabc). Then total number of arrangements on lines similar to above is 9C3 x 3C1 x (6!/2!/2! - 5!/2!) = 30240.

Fourth case : Three zeros are selected. Select other 3 digits in 9C3 ways (000abc). Number of arrangements is 9C3 x (6!/3!-5!/2!) = 5040.

Combining all the cases, summation is 294840, which is my answer to problem 1.

It seems problem 2 is solved correctly. No comment about this solution please.

Regarding problem 3

First case : 5B and 5W balls are arranged in 10 bags, one ball in one bag. one B and one W ball is left. It has three options. Option 1 : both left balls in one bag can be placed in 2 x 10!/5!/4!  = 2520 ways. Option 2 : One of the left balls can be placed in bags of bag with same type of balls and other in bags of dissimilar type of bag. It can be done in 2 x 10!/5!3! = 10080 ways. Option 3 : Both left out balls are placed in bags of similar types of bags. It can be done in 10!/4!4! = 6300 ways.

Second case : 6B and 4W balls are arranged in 10 bags. 2W balls are left. Again 3 options are there. Option 1 : Left balls of 2W placed in bags of similar typs of balls. This can be done in 2 x 10!/6!3!=1680 ways.  Option 2 : Both balls can be placed in bags containing similar types of balls. It can be done in 2 x 10!/6!/2!/2! = 2520 ways. Option 3 : Both balls are placed in bags containing disimilar types of balls. It can be done in 10!/4!4!2! = 3150 ways.

Total number of ways = 2520 + 10080 + 6300 + 1680 + 2520 + 3150 = 26250.

(Note: Multiplication by 2 for those cases, which are duplicate for B and W balls)

Re: Permutation combination
by vamsi krishna - Wednesday, 10 September 2008, 12:34 PM
  H.Sekhar>

lemme ThankU for pointing out...in prob -1 & 3

was eagerly waiting for someone to post solutions so that i can understand my weakness ..

But Prob-1 is still hazy

U Rock ON!

VaMsI
Re: Permutation combination
by ROHAN JOSHI - Friday, 3 April 2015, 03:12 PM
  Heyy Himanshu pls. explain how ...Option 1 : both left balls in one bag can be placed in 2 x 10!/5!/4! = 2520 ways

how come 10! as as you can select 1 bag out of 10 in 10 ways and not 10!
Re: Permutation combination
by Ankit Kaushik - Thursday, 19 November 2015, 02:01 AM
 

For single digit,there will be nine numbers having a single digit. For two digits, there will be two types of numbers : abbb & aabb. First let's choose two digits,it can be done in C(10,2) ways.

Now for first type abbb, arrangement of digits can be done in { 2xfac(4)/fac(3)}.Here, we have first shuffled both digits in abbb and then divided by fac(3).Also,the two digits we chose can alternately take the variable a & b.

Now,for the type aabb, we will first shuffle the digits i.e fac(4) and the divide them by fac(2) two times.(cos there are 2 a's & 2 b's).Guess why ,we are not here multiplying by 2.

So the answer will be :

9 + C(10,2) x{ [(2xfac(4)/fac(3))] + [fac(4)/(fac(2)xfac(2))] }