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Permutation and Combination Teasers II
by Rahul Jena - Sunday, 13 January 2008, 09:41 AM
  1)If each permutation of the digits 1,2,3,4,5,6 are listed in the increasing order of the magnitude, then 289th term will be:--

2)Two out of six paper sets for an examination are of mathematics.What is the number of ways in which papers can be set so that  the two mathematics papers are not together?
a) 480  b)520  c)492  d)512
 tried it by trying to find all cases where they will be together. ie 5!.
All acses are 6!.So subtracting both should give the ans.But sumwhere I am goin wrong.Plz solve in detail though I  know its a very easy question.

3)From 3 different soft drinks,4 chinese dishes and 2 ice creams, how many different  meals are possible if at least one of each of the three items is to be included , depending upon the number of people likely to turn up?
a)315  b)282  c)864  d)none of these

4)We are required to form different words with the help of the word INTEGER.Let x be the number of words in which I and N are never together and y be the number of words which begin with I and end with R,then x/y is--
a)42 b)30  c)6  d)50

5)How many integers between 1 an 100000 have sum of their digits equal to 18?
a)8993 b)25927  c)36592  d)24987

6)A difficult one :-- How many three digit numbers with distinct digits can be formed such that product of the digits is the cube of a positive integer?
a)21  b)24  c)36  d)30

7)A candidate is required to answer six out of ten  questions which are divided into groups,each containing five questions and he is not permitted to attempt more then 4 from any group.IN how many ways can he make up his choice?
a)184  b)192  c)200  d)none of these.

8)Five balls of different colors are to be placed in three boxes of different sizes.Each box can hold all five balls.In how many ways can we place the balls so that no box is empty?
a)119  b)150   c)210  d)180

9)Anil is planning to give a birthday party at his place.In how many ways can he invite one or more of his friends and seat them at a circular table?
a)84  b)89  c)78  d)81
Re: Permutation and Combination Teasers II
by TG Team - Sunday, 13 January 2008, 10:48 AM
  6)A difficult one :-- How many three digit numbers with distinct digits can be formed such that product of the digits is the cube of a positive integer?
a)21  b)24  c)36  d)30
Let one digit is fixed to be 1. Then product of two distinct digits must be a perfect cube i.e.
(2,4) such that 1*2*4 = 8 = 23 total 3! = 6 combinations
(3,9) such that 1*3*9 = 27 = 33 total 3! = 6 combinations
One digit is fixed to be 2.
(4,8) such that 2*4*8 = 64 = 43 total 3! = 6 combinations
One digit is fixed to be 3.
(8,9) such that 3*8*9 = 216 = 63 total 3! = 6 combinations
One digit is fixed to be 4
(6,9) such that 4*6*9 = 216 = 63 total 3!  = 6 combinations
Total 30 different three digit numbers can be formed with given conditions.
Re: Permutation and Combination Teasers II
by Small Wonder - Sunday, 13 January 2008, 06:00 PM
 

5) How many integers between 1 an 100000 have sum of their digits equal to 18?

a)8993 b)25927  c)36592  d)24987

a+b+c+d+e+f=18 has 23C5= 33649 integral solutions.

if a=10 then b+c+d+e+f=8 has 12C4 = 495 integral solutions.

if a=11 then b+c+d+e+f=7 has 11C4 =330 integral solutions.

if a=12 then b+c+d+e+f=6 has 10C4=210 integral solutions.

if a=13 then b+c+d+e+f=5 has 9C4=126 integral solutions.

if a=14 then b+c+d+e+f=4 has 8C4=70 integral solutions.

if a=15 then b+c+d+e+f=3 has 7C4=35 integral solutions.

if a=16 then b+c+d+e+f=2 has 6C4=15 integral solutions.

if a=17 then b+c+d+e+f=1 has 5C4=5 integral solutions.

if a=18 then b+c+d+e+f=0 has 4C4=1 integral solutions.

Total=1287 for a>10

Now, any one a, b, c, d, e, f can be more than 10.

Hence, 1287 x 6 = 7722

So, answer = 33649 - 7722 = 25927!

Small Wonder!

Re: Permutation and Combination Teasers II
by Small Wonder - Sunday, 13 January 2008, 06:03 PM
 

1)If each permutation of the digits 1,2,3,4,5,6 are listed in the increasing order of the magnitude, then 289th term will be:--

1XXXXX - 120/120

2XXXXX - 120/240

31XXXX - 24/264

32XXXX - 24/288

341256- 1/289

Small Wonder!

Re: Permutation and Combination Teasers II
by Small Wonder - Sunday, 13 January 2008, 06:08 PM
 

2) Two out of six paper sets for an examination are of mathematics.What is the number of ways in which papers can be set so that  the two mathematics papers are not together?

a) 480  b)520  c)492  d)512

X NM X NM X NM X NM X

Choose any two X out of the five Xs = 5C2 = 30

multiply this by 2! and 4! = 30 x 48 = 1440

None of these!

Small

Re: Permutation and Combination Teasers II
by Small Wonder - Sunday, 13 January 2008, 06:24 PM
 

4) We are required to form different words with the help of the word INTEGER. Let x be the number of words in which I and N are never together and y be the number of words which begin with I and end with R,then x/y is--

a)42 b)30  c)6  d)50

Number of words in which I and N are never together = total words possible - Number of words in which I and N are together

=> Number of words in which I and N are never together = 7!/2! - 6!2!/2!

number of words which begin with I and end with R = IXXXXXR = 5!/2!

ZZZZZZZzzzzzzzzzzzzzzzzzzzzzzzz!!

Small

 

Re: Permutation and Combination Teasers II
by Rahul Jena - Monday, 14 January 2008, 08:33 AM
 

Hi SMALL,

Please tell for question 2 where I am going wrong.I tried to solve it similarly the way u solved the above question.sad

Re: Permutation and Combination Teasers II
by brendan decruz - Monday, 14 January 2008, 08:36 PM
  Q3] option a} 315 ways ......
Re: Permutation and Combination Teasers II
by brendan decruz - Monday, 14 January 2008, 09:03 PM
  q7] option c] 200 ways...
Re: Permutation and Combination Teasers II
by brendan decruz - Monday, 14 January 2008, 09:07 PM
 

q9] incomplete question.....

 

Re: Permutation and Combination Teasers II
by brendan decruz - Monday, 14 January 2008, 09:31 PM
  q8] option b...150 ways...
Re: Permutation and Combination Teasers II
by Rahul Jena - Tuesday, 15 January 2008, 08:37 AM
 

Questions 9 is one or 5 ....

Brendon...plz post ur approaches......else just posting the ans is of not much use....its how its solved that important.

Also I had raised a query for secind question.....plz give a thought to it and tell me where i m going wrong...

Re: Permutation and Combination Teasers II
by Rahul Jena - Tuesday, 15 January 2008, 08:42 AM
 

smile

Hi Small...ur solutin for question 2 is wrong....The ans is 480.

I had also initially soloved like the way u had done..but sumwhere we are going wrong..........  TG SIR ...... SOS ... SOS....SOS....SOS...........

(Now only Captain Dhruv alias TG can help us..........)

Re: Permutation and Combination Teasers II
by brendan decruz - Tuesday, 15 January 2008, 04:23 PM
 

Hey Rahul,

Q2] SIMPLE APPROACH ….

Lets consider the 2 maths papers to be one entity…thus, 5 papers in all…..

No. of ways to arrange these 5 papers such that 2 maths papers occur together always…. (5! * 2!) ways

No. of ways to arrange 6 papers = 6!

Hence, no,of ways to arrange them such that the 2 papers don’t occur together

= 6! – (5!*2!)  ways

= 480 ways

Re: Permutation and Combination Teasers II
by Rahul Jena - Wednesday, 16 January 2008, 08:42 AM
 

Thanks Brendan.....U r my captain dhruv.....I was never taking that 2! into consideration.

 

Please also tell the approach for question 7 and 8

Re: Permutation and Combination Teasers II
by brendan decruz - Wednesday, 16 January 2008, 03:48 PM
 

8] look, as no box is to be empty, for box B1,B2,B3 we could have number of balls contained combinations of (3,1,1) or (2,2,1)…..for, (3,1,1) occurs in 3 ways and (2,2,1) in 3 ways.

Now, also as the balls are different b1,b2,b3,b4,b5 we have to select and then arrange them in different ways for each combination … now u can do it surely…

 

Re: Permutation and Combination Teasers II
by Mohit Soni - Monday, 16 July 2012, 07:31 PM
  for "How many integers between 1 an 100000 have sum of their digits equal to 18?

a)8993 b)25927 c)36592 d)24987"

don't you think it should be 1,000,000 instead of 1,00,000 ??
Re: Permutation and Combination Teasers II
by pranay pant - Monday, 30 September 2013, 03:14 AM
  hey shouldn't we take 5 variables instead of six cos if we take six then the nos. exceeding 1000000 will also be included. please help!
Re: Permutation and Combination Teasers II
by Rohit Gupta - Saturday, 19 October 2013, 08:45 PM
  wrong answer it should be for 1000000
Re: Permutation and Combination Teasers II
by shubham arora - Tuesday, 22 October 2013, 10:28 PM
  5c2 is 10
Re: Permutation and Combination Teasers II
by TG team - Wednesday, 6 November 2013, 04:30 PM
  3)From 3 different soft drinks,4 chinese dishes and 2 ice creams, how many different  meals are possible if at least one of each of the three items is to be included , depending upon the number of people likely to turn up?
a)315  b)282  c)864  d)none of these

No. of ways =(1 soft drink+ 2 soft drink+ 3 soft drink)(1 chinese+2 chinese+3 chinese+ 4 chinese)(1 ice cream+2 ice cream)

=3c1+3c2+3c3)(4c1+4c2+4c3+4c4)(2c1+2c2)
=315

7)A candidate is required to answer six out of ten  questions which are divided into groups,each containing five questions and he is not permitted to attempt more then 4 from any group.IN how many ways can he make up his choice?
a)184  b)192  c)200  d)none of these.

it can be done in three ways:
four from one and two from other
two from one and four from other
three from each.
(5c4*5c2)+(5c2*5c4)+(5c3*5c3)=5*10+5*10+10*10=200


8)Five balls of different colors are to be placed in three boxes of different sizes.Each box can hold all five balls.In how many ways can we place the balls so that no box is empty?
a)119  b)150   c)210  d)180

two ways
2,2,1 =3!/2!=3
3,1,1=3!/2!=3
3*(5c2*3c2*1c1) + 3*(5c3*2c1*1c1)=90+60=150




Re: Permutation and Combination Teasers II
by subhashish Bhattacherjee - Thursday, 9 July 2015, 03:09 AM
  5C2 =10