Permutations & Combinations - Questions & Discussions

MBA CAT 2022
148
sanjay04ag Says
what will be the answer if this is not a circular table but a normal linear row?


Had it not been a linear arrangement instead, the way I would have proceeded -----

Cases when 3 and 4 are always together -- 5! * 2!
Cases when (3 and 4) & (1 and 2) are together -- 4! * 2! * 2!

Required cases will be ---- (5! * 2!) - (4! * 2! * 2!)

Sayonara
149
4 balls are to be put in 5 boxes.in how many ways can this be done if
a)Balls are similar and boxes are different
This is similar to finding the whole number solutions for the equation
a + b + c + d + e = 4
which is given by (n+r-1) C r-1 where n is the constant on the RHS and r is the number of variables.
Hence in this case we have 8 C 4 = 70 ways

c)both boxes and balls are similar
Here its the number of ways you can split 4 into 5 numbers
40000
31000
22000
21100
11110
Hence 5 ways

b)balls are different and boxes are similar
For each of the cases listed in (c) the number of ways of selecting the balls has to be considered
For eg, for the case 40000, the 4 balls for the first box can be selected in only 1 way.
For the case 31000 the 3 balls for the first box can be selected in 4C3 = 4 ways
In 22000 2 balls for the first box can be selected in 4C2 = 6 ways (note the remaining 2 balls automaticallly select themselves for the second box). Similarly in 21100 the 2 balls to go together can be selected in 6 ways, the remaining two balls automatically go into different boxes.Finally, in 11110 there is just 1 way of splitting all the balls into 4 boxes. Hence, total number of ways = 1+4+6+6+1 = 18 ways

Note: The number of positive integral solution(natural number) for
x1+x2+x3+.........xr=n is (n-1) C (r-1)


Hi .........
I am still not fully clear with CASE B:
Can you plz explain step by step the number of ways when 6 different balls are placed in 7 identical boxes?

the cases that arise would be:
(6,0,0,0,0,0,0)
(5,1,0,0,0,0,0)
(4,2,0,0,0,0,0)
(4,1,1,0,0,0,0)
(3,3,0,0,0,0,0)
(3,2,1,0,0,0,0)
(3,1,1,1,0,0,0)
(2,2,2,0,0,0,0)
(2,2,1,1,0,0,0)
(2,1,1,1,1,0,0)
(1,1,1,1,1,1,0)

Can you please explain how the combinations of each case are formed?
150

Can someone post the approach for the four cases of Permutations and Combinations i.e., 1) Similar ti Similar, 2) Different to similar, 3) Similar to Similar and 4)Different to Different. Even links to these cases would help!!
Thanks in Advance! 😃

151

#1 There are 5 boxes and 5 balls, in how many ways can all the balls be put in boxes if,

a) boxes and balls are all distinct and only 1 ball in one box
b)boxes are same and balls are distict and only 1 ball per box
c)balls are same and boxes are different and only 1 ball per box
d)balls and boxes are same and only 1 ball per box
e)repeat all above questions with any number of balls in any box

152
Hi .........
I am still not fully clear with CASE B:
Can you plz explain step by step the number of ways when 6 different balls are placed in 7 identical boxes?

the cases that arise would be:
(6,0,0,0,0,0,0)
(5,1,0,0,0,0,0)
(4,2,0,0,0,0,0)
(4,1,1,0,0,0,0)
(3,3,0,0,0,0,0)
(3,2,1,0,0,0,0)
(3,1,1,1,0,0,0)
(2,2,2,0,0,0,0)
(2,2,1,1,0,0,0)
(2,1,1,1,1,0,0)
(1,1,1,1,1,1,0)

Can you please explain how the combinations of each case are formed?

Can someone plz help out with this prob. that was quoted in my last post as I am still unable to decipher the way to solve this?
153
swaminathan21 Says
Can someone plz help out with this prob. that was quoted in my last post as I am still unable to decipher the way to solve this?


Balls different but boxes similar. This means that we need the "combination equation" to select the balls, but the boxes select themselves.

Case 1: (6-0-0-0-0-0-0). There is only one way to select 6 balls out of 6. No. of ways for this selection = 1.

Case 2: (5-1-0-0-0-0-0). Select 5 balls out of 6 in 6 ways. The other ball goes into another box automatically. No. of ways for this selection = 6.

Case 3: (4-2-0-0-0-0-0). Select 4 balls out of 6 in C(6,4) = 15 ways. Remaining 2 balls are selected automatically. No. of ways for this selection = 15.

Case 4: (4-1-1-0-0-0-0). Select 4 balls in 15 ways. For the other 2 balls, they automatically go into different boxes. There is no need to apply combination equation as the boxes are same, so there is no need to select them especially. No. of ways for this selection = 15.

Case 5: (3-3-0-0-0-0-0). Select 3 balls from 6 in C(6,3) = 20 ways. There is no need to select remaining balls as they are left out, so get selected automatically. No. of ways for this selection = 20.

Case 6: (3-2-1-0-0-0-0). Select 3 balls in 20 ways. Of the remaining 3 balls, we need to select 2 balls for one box. This selection can be made in C(3,2) = 3 ways. Total no. of ways to make this selection = 20 X 3 = 60.

Case 7: (3-1-1-1-0-0-0). 20 ways. Here, the three balls go into different boxes automatically.

Case 8: (2-2-2-0-0-0-0). Select 2 balls from 6 in C(6,2) = 15 ways. Of the remaining 4 balls we need to select 2 balls for a box, in C(4,2) = 6 ways. The other 2 balls automatically into other box. Total no. of ways to make this selection = 15 X 6 = 90.

Case 9: (2-2-1-1-0-0-0). Select 2 balls of 6 in 15 ways. Then select 2 more balls from remaining 4 in 6 ways. Last 2 balls will go into different boxes. We do not have to select them as the boxes are same. Total no. of ways for this selection = 15 X 6 = 90.

Case 10: (2-1-1-1-1-0-0). Only 15 ways, as the remaining balls go into different boxes on their own.

Case 11: (1-1-1-1-1-1-0). There is only 1 way to put all the balls in different boxes! As the boxes are same, we do not have to select the empty box.

Total no. of ways = 1+6+15+15+20+60+20+90+90+15+1 = 333 ways to put the balls in 7 boxes.
154

Thanks CRAB!!!
Thy hath dispelled my doubts.
Hope the question wud have brought in much clarity to distribution concept in P&C;.

155
Balls different but boxes similar. This means that we need the "combination equation" to select the balls, but the boxes select themselves.

Case 1: (6-0-0-0-0-0-0). There is only one way to select 6 balls out of 6. No. of ways for this selection = 1.

Case 2: (5-1-0-0-0-0-0). Select 5 balls out of 6 in 6 ways. The other ball goes into another box automatically. No. of ways for this selection = 6.

Case 3: (4-2-0-0-0-0-0). Select 4 balls out of 6 in C(6,4) = 15 ways. Remaining 2 balls are selected automatically. No. of ways for this selection = 15.

Case 4: (4-1-1-0-0-0-0). Select 4 balls in 15 ways. For the other 2 balls, they automatically go into different boxes. There is no need to apply combination equation as the boxes are same, so there is no need to select them especially. No. of ways for this selection = 15.

Case 5: (3-3-0-0-0-0-0). Select 3 balls from 6 in C(6,3) = 20 ways. There is no need to select remaining balls as they are left out, so get selected automatically. No. of ways for this selection = 20.

Case 6: (3-2-1-0-0-0-0). Select 3 balls in 20 ways. Of the remaining 3 balls, we need to select 2 balls for one box. This selection can be made in C(3,2) = 3 ways. Total no. of ways to make this selection = 20 X 3 = 60.

Case 7: (3-1-1-1-0-0-0). 20 ways. Here, the three balls go into different boxes automatically.

Case 8: (2-2-2-0-0-0-0). Select 2 balls from 6 in C(6,2) = 15 ways. Of the remaining 4 balls we need to select 2 balls for a box, in C(4,2) = 6 ways. The other 2 balls automatically into other box. Total no. of ways to make this selection = 15 X 6 = 90.
Case 9: (2-2-1-1-0-0-0). Select 2 balls of 6 in 15 ways. Then select 2 more balls from remaining 4 in 6 ways. Last 2 balls will go into different boxes. We do not have to select them as the boxes are same. Total no. of ways for this selection = 15 X 6 = 90.

Case 10: (2-1-1-1-1-0-0). Only 15 ways, as the remaining balls go into different boxes on their own.

Case 11: (1-1-1-1-1-1-0). There is only 1 way to put all the balls in different boxes! As the boxes are same, we do not have to select the empty box.

Total no. of ways = 1+6+15+15+20+60+20+90+90+15+1 = 333 ways to put the balls in 7 boxes.


Everything seems to be fine except the three cases (in bold)

(3, 3, 0, 0, 0, 0, 0) - C(6, 3) = 20, but total number of cases will be 10 as (abc, def) and (def, abc) are same

(2, 2, 2, 0, 0, 0, 0) - C(6, 4)*C(4, 2) = 90, but total cases will be 15, as here order of choosing the pair is not important so we need to divide by 3! = 6.

(2, 2, 1, 1, 0, 0, 0) - C(6, 2)*C(6, 2) = 90, but total cases will be 45, as here also the order is not important, so we need to divide by 2!

So, the answer should be 203
156
Everything seems to be fine except the three cases (in bold)

(3, 3, 0, 0, 0, 0, 0) - C(6, 3) = 20, but total number of cases will be 10 as (abc, def) and (def, abc) are same

(2, 2, 2, 0, 0, 0, 0) - C(6, 4)*C(4, 2) = 90, but total cases will be 15, as here order of choosing the pair is not important so we need to divide by 3! = 6.

(2, 2, 1, 1, 0, 0, 0) - C(6, 2)*C(6, 2) = 90, but total cases will be 45, as here also the order is not important, so we need to divide by 2!

So, the answer should be 203



Noted...thanks for taking out the time and pointing it out...
157
Everything seems to be fine except the three cases (in bold)

(3, 3, 0, 0, 0, 0, 0) - C(6, 3) = 20, but total number of cases will be 10 as (abc, def) and (def, abc) are same

(2, 2, 2, 0, 0, 0, 0) - C(6, 4)*C(4, 2) = 90, but total cases will be 15, as here order of choosing the pair is not important so we need to divide by 3! = 6.

(2, 2, 1, 1, 0, 0, 0) - C(6, 2)*C(6, 2) = 90, but total cases will be 45, as here also the order is not important, so we need to divide by 2!

So, the answer should be 203


Hi CHILL Bhai and Crab,
Consider the case:(3, 3, 0, 0, 0, 0, 0).
The cases that wud arise with a,b,c,d,e and f balls are:
1.abc
2.abd
3.abe
4.abf
5.acd
6.ace
7.acf
8.ade
9.adf
10.aef
11.bcd
12.bce
13.bcf
14.bde
15.bdf
16.bef
17.cde
18.cdf
19.cef
20.def.

So in all 20cases of grouping into 3.
Case1 will complement case20...similarly (2,19)...(3,18 )etc...In all 10 WAYS.

Can I have clarity on when u will divide by 2!,3! etc.?
Is there any generalised method of solving this?
158
Hi CHILL Bhai and Crab,
Consider the case:(3, 3, 0, 0, 0, 0, 0).
The cases that wud arise with a,b,c,d,e and f balls are:
1.abc
2.abd
3.abe
4.abf
5.acd
6.ace
7.acf
8.ade
9.adf
10.aef
11.bcd
12.bce
13.bcf
14.cde
15.cdf
16.def.

So in all 16 cases of grouping into 3.
I know this is wrong. But this is what i got when grouping.
To me crab's method sounded fine.
Can I have clarity on when u will divide by 3! etc.?
Is there any generalised method of solving this?


I think you missed following 4 cases:-
bde, bdf, bef, cef

Seems you didn't get the point
(3, 3, 0, 0, 0, 0, 0) - If first 3 is for abc, then second 3 has to be for def.
Now, in 20 you are considering a case when first three is for def and second one is for abc. But both these cases are same as boxes are same, so order doesn't matter here.

I hope it helps, if its not then you can PM me.
159

I have edited that. Sorry to have missed it.

160

Thanks, much appreciated 😃

161

what is the difference between the Permutations and Combinations ?

162
Permutations is simply the arrangement of certain objects differently.
Remember here that the order in which you select the items is important.
Ex: Suppose you have to find out the count of all the 5 digit numbers possible from digits 1,2,3,4,5 when repition of digits is not allowed.
The answer is : 5*4*3*2*1 = 5!
Reason: Suppose the folowing are the five blanks:
_ _ _ _ _.
* You can fill the first blank in 5 ways(Select any digit from 1-5)
* You can fill the second blank in 4 ways(Select any digit from 1-5 except the one already selected in the first blank as repitition of digits is not allowed.)
* You can fill the third blank in 3 ways(Select any of the 3 numbers left from 5)
Similarly for the 4th and 5th blank.

-If the condition would have been changed so that repition of digits is allowed, then the number of possibilities would have been:
5*5*5*5*5
This is because you can fill all the blanks by any of the numbers possible.

Suppose you have to arrange A,B,and C in all possible ways,without repition, you can do that in the following ways:

AB,AC,BA,BC,CA,CB

As you can notice AB is different from BA,AC is different from CA and so on.
So, there are a total of 6 ways.

The method fo this is: The first blank can be filled in 3 ways(from A,B, and C) and the second can be filled in 2 ways(from the left over alphabets)
This is the basic concept of permutation which is further modified to give different results.

Combinations:

Combinations are the number of ways of selecting a few things from a certain number of things.
Remember here the order in which you select the things is not important.
Ex: Suppose you have to count the number of different teams possible each having 3 members each from a set of 5 players.

Let the set of players be A,B,C,D,E
The combinations can be:
ABC,ABD,ABE,ACD,ACE,ADE,BCD,BCE,BDE and CDE.

Remember here ABC is not different from CAB because what you are looking at is combination. Both ABC and CAB are same because there are the same set of players involved in both the teams.

This is the main difference between permutations and combinations.

You can find the short method to calculate the number of permutations and combinations from any book. This is the basic funda of permutations and combinations. So, I am going to stop here.
Hope this helped.
163

please help me out with the following question.
1) how many new ways are there to choose four cards of different suits and different values from a deck of 52 card.(Please do expain your approach)
a. 13*12*11*10
b.52C4
c. 13 raise to power 4
d. 52*36*22*10

164

how many words are possible from the letters of the word PERMUTATION?
a. 11!/2!
b. (11!/2!)-1
c. 11!-1

(please do explain the approach)

165
please help me out with the following question.
1) how many new ways are there to choose four cards of different suits and different values from a deck of 52 card.(Please do expain your approach)
a. 13*12*11*10
b.52C4
c. 13 raise to power 4
d. 52*36*22*10


Answer shud be option d.
ie, 52*36*22*10

First the cards set can be divided into 4...spade,clubs,hearts and dice.
first out of 4, i choose 1 set.
4c1.
from that selected set, there are 13 cards.out of those 13 cards , i choose 1 card.
=13c1.
so selecting 1 card from that selected set=4c1*13c1=52.---->(1)

now from the remaining 3 sets i choose 1 set=3c1.
but here u cannot select from the 13 cards. u can select only from 12 cards, as u cannot select the card selected in set 1.
so in all 3c1*12c1=36.------------------------------>(2)

now remaining are 2 sets.
so no.of ways of selecting 1 set from the 2 sets =2c1.
in this set u can select only 11 cards, as u cannot select previously selected 2 cards in (1)&(2).
so in all 2c1*11c1=22----------------------->(3)

now the remaining set gets automatically selected. from this u select 1 card out of 10 selectable cards.
so 10C1=10----------------------------------------->(4).

final answer is (1)and(2)and(3)and(4). And symbolises *(multiplication).
so 52*36*22*10 is the required answer.
166
how many words are possible from the letters of the word PERMUTATION?
a. 11!/2!
b. (11!/2!)-1
c. 11!-1

(please do explain the approach)


answer option a.

no.of letters in permutation=11.
so no. of words tat can be formed=11!.
but there are 2 t's . so divide by 2!.

final answer=11!/2!.

if no repeating alphabets...dont divide.
if alphabet repeats n times, divide by n!.
167

Thanks Swaminathan, and wonderfull explaination too. please help me with the other question and the following question.


There are 5 blue socks, 4 red socks and 3 green socks in Debu's wardrobe. He has to select 4 socks from this set. In how many ways can he do so?

A. 245
B. 120
C. 495
D. 60