Permutations & Combinations - Questions & Discussions

MBA CAT 2022
309
spectramind07 Says
Buddy you solved a question that was posted in 2007!!!!:w00t:

Bro i m just going thru all the permutation and combination problem posted in pg as i had ignore it all thru my life(actually i ignore everything). Its just to get taste of different question types
310
Puys I hv a doubt in the following Qn
In how many ways can a person distribute greeting cards (obviously not identical ones) to 6 different friends when he has 4 servants to do so.
The answer is 4^6
Explaination every cards has 4 options to chose from so 4 x 4 x 4 x 4 x 4 x 4= 4^6
My doubt is : why cant we go the other way around namely every servant has
6C0 + 6C1 + 6C2 + 6C3 + 6C4 + 6C5 + 6C6 = 2^6 options
now there r 4 servants so 4 x 2^6 = 4^4 which is incorrect.
Can someone plz point out the error. Im eagerly waiting for someone to clarify it. Thank u.

First greeting card can be taken by 4 servant,
Second greeting card can be taken by 4 servant..
and so on........
considering ur case,
First servant can carry 6 greeting cards.
Second servant can carry 6 greeting card(this is wrong as they are dependent event, and so is your doubt)
311
spectramind07 Says
Buddy you solved a question that was posted in 2007!!!!:w00t:

Just doing some stupid parts of Qa which i will never do from book
312
Hi puys

Please help me with this :


Question : 8 Similar thing to be distributed among 4 different students ?

Answer : 11! / 8! 3!.
======================================

Now when the question iis :

Question : 8 Different things to be distributed among 4 different students?

Answer : 11! / 3! OR 4^8 ?????????
================================

Thanks
Deepak


Just look at these basic formulae of permutation and combination and u will b done

No of ways of dividing n distinct object among r distinct grps = r^n
Your second question

No of ways of dividing n distinct object into r identical grps(grp can be empty)=r^n/r!
No of ways of dividing n identical object among r distinct grps(grp can be empty)=n+r-1cr-1
No of ways of dividing n identical object among r distinct grps(grp can't be empty)=n-1cr-1..
hope its clr
313
as there is no restriction given on selection there are 4 ways of selecting each disk. hence i think answer should be 4^6.

please correct me if i'm wrong.

Ya bro, u are wrng..you are taking it in this manner,
first disk can be slected in 4 ways,
second disk can be slected in 4 ways..and so on
but this we can take only if all disk being distinct which is not the case here,

Pls look at the post above to get the forumla, if u want proof, pm me
314

Doubts :

1.The number of terms in the expansion:

(a+b+c+d)^20 ............Plz don tell the formula oly Plz explain also

Ans:1771

2.In how many ways can 4 prizes each havin 1 2nd and 3rd positions be given to 3 boyss if each boy is elig to reci one prize for each event?

Ans:6^4

3.In how many ways can 15 books be divided equally?

(1) among 3 boyss?

Ans: fact 15/(fact5)^3

(2). Into 3 parcels?

Ans:fact 15/(fact 3 * (fact 5 )^3))

4.Find d sum of numbers that can be formed by taking all the digits at the time from 4,5,6,7,8 without repetion??


Ans:7999929


5.The number of non neg integral solutions for

x+y+z+t = 5 is

Ans:56

Plz solve

315
Doubts :

2.In how many ways can 4 prizes each havin 1 2nd and 3rd positions be given to 3 boyss if each boy is elig to reci one prize for each event?

Ans:6^4


If you consider the case of 1 game (1 prize in your question) which has positions 1,2 and 3.

Then you have 3 boys and 3 possible positions. So the number of permutations is 3P3 => 3! = 6 ways

Since the 4 games (4 prizes) are all unique with the same 3 boys. Each game can be done in 3P3 ways.

Thus you have (no of ways game 1 can be won) * (no of ways game 2 can be won) *(no of ways game 3 can be won) *(no of ways game 4 can be won)

6*6*6*6 = 6^4


Doubts :

3.In how many ways can 15 books be divided equally?

(1) among 3 boyss?

Ans: fact 15/(fact5)^3

(2). Into 3 parcels?

Ans:fact 15/(fact 3 * (fact 5 )^3))



The first boy will have 15 books to choose 5 books from. Thus 15C5
The second boy can select 5 books from the remaining 10 books. Thus 10C5
The third boy picks up the remaining 5 books. 5C5

15C5 * 10C5 * 5C5
(15! / 5! 10!) * (10! / 5! 5!) * (1)

Hence 15! / (5!)^3

In the first case the groups were unique (3 different boys). In the second case the groups are the same (the parcels are identical) => Earlier the groups were ABC. Now they are AAA. Thus from the law of repetition, we need to divide by the factorial of the number of repetitions that happened.

Hence / 3!


Doubts :

5.The number of non neg integral solutions for

x+y+z+t = 5 is

Ans:56


The integral solutions are as follows.
5 0 0 0
4 1 0 0
3 2 0 0
3 1 1 0
2 2 1 0
2 1 1 1

These solutions can be distributed over the 4 variables respectively as

4! / 3! = 4
4! / 2! = 12
4! / 2! = 12
4! / 2! = 12
4! / 2! = 12
4! / 3! = 4

Summing this up gives us 56.
316
If you consider the case of 1 game (1 prize in your question) which has positions 1,2 and 3.

Then you have 3 boys and 3 possible positions. So the number of permutations is 3P3 => 3! = 6 ways

Since the 4 games (4 prizes) are all unique with the same 3 boys. Each game can be done in 3P3 ways.

Thus you have (no of ways game 1 can be won) * (no of ways game 2 can be won) *(no of ways game 3 can be won) *(no of ways game 4 can be won)

6*6*6*6 = 6^4




The first boy will have 15 books to choose 5 books from. Thus 15C5
The second boy can select 5 books from the remaining 10 books. Thus 10C5
The third boy picks up the remaining 5 books. 5C5

15C5 * 10C5 * 5C5
(15! / 5! 10!) * (10! / 5! 5!) * (1)

Hence 15! / (5!)^3

In the first case the groups were unique (3 different boys). In the second case the groups are the same (the parcels are identical) => Earlier the groups were ABC. Now they are AAA. Thus from the law of repetition, we need to divide by the factorial of the number of repetitions that happened.

Hence / 3!




The integral solutions are as follows.
5 0 0 0
4 1 0 0
3 2 0 0
3 1 1 0
2 2 1 0
2 1 1 1

These solutions can be distributed over the 4 variables respectively as

4! / 3! = 4
4! / 2! = 12
4! / 2! = 12
4! / 2! = 12
4! / 2! = 12
4! / 3! = 4

Summing this up gives us 56.


Thx ............solutions for the remaining 2 questions plzzz
317
Doubts :


4.Find d sum of numbers that can be formed by taking all the digits at the time from 4,5,6,7,8 without repetion??

Ans:7999929

Plz solve



Say, 4 is at left most posistion.
Remaining positions can be taken in 4! ways.
So, 4 occupies left most position 24 times.
Similarly, all numbers occupy each position 24 times.

If 4 occurs 24 times at unit place, sum = 24*4*1
If 4 occurs 24 times at 10s place, sum = 24*4*10
If 4 occurs 24 times at 100s place, sum = 24*4*100
If 4 occurs 24 times at 1000s place, sum = 24*4*1000
If 4 occurs 24 times at 10000s place, sum = 24*4*10000

So, sum for all 4s = 24*4*(11111)

So, sum for all numbers = 24*11111*(4+5+6+7+8 ) = 7999920
318
Say, 4 is at left most posistion.
Remaining positions can be taken in 4! ways.
So, 4 occupies left most position 24 times.
Similarly, all numbers occupy each position 24 times.

So, sum = 24*11111*(4+5+6+7+8 ) = 7999920


Doubt in bold, got ur above steps
319
mbafrmfms Says
Doubt in bold, got ur above steps


Hi,

I have now added details in original post. Please refer to that one.
320
Say, 4 is at left most posistion.
Remaining positions can be taken in 4! ways.
So, 4 occupies left most position 24 times.
Similarly, all numbers occupy each position 24 times.

If 4 occurs 24 times at unit place, sum = 24*4*1
If 4 occurs 24 times at 10s place, sum = 24*4*10
If 4 occurs 24 times at 100s place, sum = 24*4*100
If 4 occurs 24 times at 1000s place, sum = 24*4*1000
If 4 occurs 24 times at 10000s place, sum = 24*4*10000

So, sum for all 4s = 24*4*(11111)

So, sum for all numbers = 24*11111*(4+5+6+7+8 ) = 7999920


I have two questions on this.

a) When we consider 4 occurring in the units place, 5 at some point of time will occur in the tens place. ie XXX54. (Where X={6,7,8} )

When we replicate the process outlined in your solution for 5 in the tens place we have XXX5X (where X={4,6,7,8}). In some of these cases 4 will be in the units place that is XXX54 (where X={6,7,8}).

Thus wont the above method of counting result in a double count?

b) The result was supposed to be 7999929. We ended up with 7999920. Was this a typo in the original question?
321

For the sum, mine is a longer yet a more convenient process. Lets start with the last digit and move leftwards.
we know that the last digit can be either of the 5 numbers 4! times. Therefore 4!(4+5+6+7+=720
The last digit is 0 carry 72. Now the same digits will be repeated in the tens column. Therefore 720+72=792.Take 2 carry 79.
We know the lasst 2 digits as 2 and 0. If we have the options we can easily put it there. Else go on 720+79=799.Take 9 and carry 79.Now 79 will be repeated. Therefore the number will be 7999920.

322
I have two questions on this.

a) When we consider 4 occurring in the units place, 5 at some point of time will occur in the tens place. ie XXX54. (Where X={6,7,8} )

When we replicate the process outlined in your solution for 5 in the tens place we have XXX5X (where X={4,6,7,8}). In some of these cases 4 will be in the units place that is XXX54 (where X={6,7,8}).

Thus wont the above method of counting result in a double count?

b) The result was supposed to be 7999929. We ended up with 7999920. Was this a typo in the original question?


I will answer second question first, because it is easy
And in CAT, we should try to answer all easy questions
Yeah.. I feel there is a typo in original question.

Now, coming to first part. See.. it goes like this:
We have 4 in unit place 24 times.
So, it will 4+4+4+4... 24 times. So, 24*4. Here it does not matter what is there at 10s or 100s place.
Similarly, 24 times we have 4 in 10s places.
So, 40+40+40+40... 24 times = 24*40
And so on.
See... if a number is 12345; it is like 10000+2000+300+40+5

So, sum of all 4s in number will be 24*(40000+4000+400+40+4)
= 24*4*11111

Similary for other numbers.
So, finally we will get 24*(2+3+4+5+6)*11111
323
2) 3*2^7 - 3 = 384 - 3 = 381

3*2^7 are the cases when atleast one of them haven't got anything, but few cases are repeated since here all those cases are also there when two of them haven't got anything

3 are the cases when only one of them got all objects
_______________________________________________________________

3) Three cases:-

i) (1, 1, 5) - 3*C(7, 1)*C(6, 1) = 126 ways
3 because 5 can be given to any one of the three

ii) (1, 2, 4) - 6*C(7, 1)*C(6, 2) = 630
6 because (1, 2, 4) can be permuted in 3! ways for three persons

iii) (1, 3, 3) - 3*C(7, 1)*C(6, 3) = 420
3 because 1 can be given to any one of the three

Total = 126 + 630 + 420 = 1176


Hi Chillfactor,

I have one doubt in this sum .. is it assumed that no person gets 0 objects .. (1,0,6) and (1,6,0) not taken into consideration ?
324
Hi Chillfactor,

I have one doubt in this sum .. is it assumed that no person gets 0 objects .. (1,0,6) and (1,6,0) not taken into consideration ?


As per the language of the question, this case should be included.

(1, 0, 6) - 6*C(7, 1) = 42 cases

So, answer should be 1218 or it should have been mentioned that they all gets at least one.
325

The AMS MOCK CAT test CATALYST 19 consists of 4 sections. Each section has a maximum of 45 marks. Find the number of ways in which a student can qualify in the AMS MOCK CAT if the qualifying marks is 90.

1) 36546
2) 6296
3) 64906
4) n.o.t

326
The AMS MOCK CAT test CATALYST 19 consists of 4 sections. Each section has a maximum of 45 marks. Find the number of ways in which a student can qualify in the AMS MOCK CAT if the qualifying marks is 90.

1) 36546
2) 6296
3) 64906
4) n.o.t


Students score in each section is a, b, c, d, then

a + b + c + d = 90
=> C(93, 90) ways, but we have counted those cases also when a, b, c, d > 45

When a > 45, i.e., a = 46 + a'
=> a' + b + c + d = 44
=> C(47, 44) ways

Similarly for other variables

So, total number of ways = C(93, 90) - 4*C(47, 44) = 64096
327

Tussi great ho sir jee!! kya accuracy hai!!

sorry for the spam, cudnt resist 😛

328

Find the no of ways in which 10 different balls can be placed in 4 different boxes such that at max 2 boxes are empty..