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Questions on Permutations and Combinations.....plz help!!!
by Nakul Kothari - Tuesday, 18 August 2009, 10:12 AM
 

Please provide detailed approach of solving these problems...

Q 1. In an exam the max marks for each of three papers is 50 each. The max marks for the fourth paper is 100. Find the no of ways with which a student can score 60% marks in aggregate?

                                                                           Given Ans.:-1,10,551 

Q 2. Seven diff objects must be divided among three people. Find the no. of ways:-

1. If one or two of them must get no objects.            Ans.:-381

2. If atleast one of them gets exactly one object.       Ans.:-1176

 

Q 3. How many no. smaller than 2.108   and divisible by 3 can be written by means of digit 0,1 and 2(exclude the single and double digit numbers)?

                                                                          Ans.:- 4373 

Re: Questions on Permutations and Combinations.....plz help!!!
by Nakul Kothari - Friday, 21 August 2009, 03:13 PM
  I again scratched my head for a long time but of no use...please guys help me to solve these questions!!!!
Re: Questions on Permutations and Combinations.....plz help!!!
by Himanshu Shekhar - Saturday, 22 August 2009, 12:08 PM
 

Only question 2 is solved.

Question no 1: No shortcut method is evolving. I am not able to figure out. Number is too big to build up through considering various cases.

Question no 2: Let three persons are getting x1, x2 and x3 objects. We have to search solution x1 + x2 + x3 = 7. As all are different objects and persons are also distinct, direct application of permutation & combination will not help.

For part 1 of Q2: Four cases of distribution are there. Case I - (0,0,7) can be done in 3 ways. Case II - (0,1,6) can be done in 7C1 x 3! = 42. Case III - (0,2,5) can be done in 7C2 x 3! = 126. Case IV - (0,3,4) can be done in 7C3 x 3! = 210. Total number of cases = 3 + 42 + 126 + 210 = 381.

For part 2 of Q2: Three cases are considered. Case I - (1,1,5) can be done 7C16C1 x 3 = 126. Case II - (1,2,4) can be done in 7C16C3 x 3! = 630. Case III - (1,3,3) can be done in 7C1 x 6C3 x 3 = 420. Total number of cases = 126 + 630 + 420 = 1176. 

Question 3: Not able to understand the problem. 2.108 is 2108 or 2x108 or 2x10108.

Himanshu Shekhar

Re: Questions on Permutations and Combinations.....plz help!!!
by Nakul Kothari - Saturday, 22 August 2009, 03:13 PM
  Thanks Himanshu
   I was also not jealous enough to take the large number of cases in Q1. As for Q2 I was making a minor mistake which i have corrected now.
In Q3 the number is 2 x 108 ......I am having no clue of solving this question....
Re: Questions on Permutations and Combinations.....plz help!!!
by umesh singh - Saturday, 22 August 2009, 11:59 PM
 

hi himanshu,

 Brother you missed one case in part 2 solution of question 2

that is (1,0,6)

umesh

Re: Questions on Permutations and Combinations.....plz help!!!
by Nakul Kothari - Monday, 24 August 2009, 10:11 AM
  @ himanshu
hey can you please explain how you have taken the permutationa in various cases of Q 2 . Like for case (0,1,6) how you have taken it to 7C1 X 3!
please explain....i think am missing out some thing
Re: Questions on Permutations and Combinations.....plz help!!!
by Himanshu Shekhar - Sunday, 30 August 2009, 07:59 AM
 

Ans 3: 4371 (if single and double digit numbers are excluded) and 4373 (if they are included).

My approach: Count numbers. Obviously we are looking for a maximum nine digit number. Approach is to find total numbers possible with certain number of digits first. Since every third number is divisible by 3, divide total number of certain-digit number by 3. Different cases are explained below.

Case 1: 9-digit numbers - Total numbers - 1 x 38; Requirement = 1 x 37 = 2187.

Case 2: 8-digit numbers - Total numbers - 2 x 37; Requirement = 2 x 36 = 1458.

Case 3: 7-digit numbers - Total numbers - 2 x 36; Requirement = 2 x 35 = 486.

Case 4: 6-digit numbers - Total numbers - 2 x 35; Requirement = 2 x 34 = 162.

Case 5 : 5-digit numbers - Total numbers - 2 x 34; Requirement = 2 x 33 = 54.

Case 6 : 4-digit numbers - Total numbers - 2 x 33; Requirement = 2 x 32 = 18.

Case 7 : 3-digit numbers - Total numbers - 2 x 32; Requirement = 2 x 31 = 6.

Case 8 (Optinal) : 2-digit numbers - Total numbers - 2 x 31; Requirement = 2.

Case 9 (Optional) : 1-digit numbers - Total numbers - 2; Requirements = 0.

Total : 2187+1458+486+162+54+18+6 = 4371 (Excluding single and double digit number)

or 4371+2 = 4373 (including single and double digit numbers).

Please check for calculation mistakes, if any.

Himanshu Shekhar

 

Re: Questions on Permutations and Combinations.....plz help!!!
by deepak nayak - Monday, 31 August 2009, 03:31 PM
  Ans 1.
In 1st three papers he can get marks 0,1,2,.......,50 and in fourth paper he can get 0,1,2,...........,100. In order to get 60% aggregate he needs 150 marks out of 250 total.

Now, see carefully...

if he get x1, x2, x3 and x4 marks in four papers then, the no. of ways to get 150 marks is same as the no. of integral soln of this eqn:
                x1 + x2 + x3 + x4 = 150
with constraints,  0<= x1, x2, x3 <= 50
                       and  0<= x4 <= 100.

Again, this no. of soln will be equal to the co-efficient of x150 in the expression:
(x0+ x1+ .....+x50).(x0+ x1+ .....+x50).(x0+ x1+ .....+x50).(x0+ x1+ .....+x100)

If, this is clear then the rest is just calculation.
so we can write above exp as:
=> (x0+ x1+ .....+x50)3.(x0+ x1+ .....+x100)
=>(1-x51)3(1-x101)3(1-x)-4

=>(1-3x51-x101+3x102+ . .).(1-x)-4

I hope, now you can solve it. I could have given you the whole solution, but I am having a hell of time typing mathematical expressions. I don't know how I can type factorial something upon something type expression.


After solving you'll get,
=>585276 - 515100 - 22100 + 62475
=>110551

Re: Questions on Permutations and Combinations.....plz help!!!
by Himanshu Shekhar - Monday, 31 August 2009, 10:01 PM
 

Answer to Umesh please.

Question reproduced - If atleast one of them gets exactly one object.   

My interpretation - one is getting exactly one object and others getting more than one object.

Himanshu Shekhar

Re: Questions on Permutations and Combinations.....plz help!!!
by Himanshu Shekhar - Monday, 31 August 2009, 10:08 PM
 

Answer to Nakul please

For case (0,1,6) : OUt of 7 available objects, we have to make three groups containing 0,1 and 6 objects. If I select 1 object out of 7, rest I can place as a group and one null group is created. This can be done in 7C1 ways. 3 groups of objects can be allotted to 3 persons in 3! ways. So answer is 7C1 x 3!.

Himanshu Shekhar

Re: Questions on Permutations and Combinations.....plz help!!!
by soudip sanyal - Monday, 2 March 2015, 01:17 PM
  Here you've written ->
For part 2 of Q2: Case II - (1,2,4) can be done in 7C1 x 6C3 x 3! = 630.
but I think it would be, 7C1 x 6C2 x 3! = 630.
smile thanx by the way smile