 Permutations & Combinations - Questions & Discussions

.Can someone help me out with the following problems with some clear and simple approcah to these qustions 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 b)balls are …

.Can someone help me out with the following problems with some clear and simple approcah to these qustions

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
b)balls are different and boxes are similar
c)both boxes and balls are similar
in which of the above cases the formulae (n+r-1)Cr-1 be applied?
is the formulae (n+r-1)Cr or (n+r_1)Cr-1
and which of these formuale is the number of positive integral soultion for
x1+x2+x3+.........xr=n

a)
as all 4 balls are same so if we arrnage them among them selves then onli one way , let me explain it further
lets take 4 rs1 coins to b given to 4 kids n each kid shd b given 1 rs exactly , there is only one way u can give , it doesnt matter which coin u give to which boy , end result will b same as all coins are identical .

we can arrange these 4 IDENTICAL balls in 5 DIFF boxs in 5x4x3x2
u can derive this very simply lik this : 1st red ball will hv 5 boxes to choose , or in other words we can put 1st red ball in any of 5 diff boxes, so wen1st red ball takes 1 box then 2nd red ball can b put in ny of remaining 4 diff bxes , simlilarly 3 rd ball can b put in ny of remaining 3 boxes , 4th ball in ny of remaining 2 boxes
so we get 5x4x3x2
we put x( multiplication sign ) in between 5 4 3 n 2 bc we r using word AND
we will read e solution lik : we can put 1st red ball in any of 5 diff boxes AND 2nd red ball can b put in ny of remaining 4 diff bxes AND 3 rd ball can b put in ny of remaining 3 boxes AND 4th ball in ny of remaining 2 boxes
so ans is 5x4x3x2x1( this 1 is for 1 way of arranging balls )

b)
4 diff balls n 5 same boxes , so in this case boxes wont matter , balls will
we can choose 4 identical boxes out of 5 identical boxes in 5c4 ( if there were only 4 identical boxes n we hv to choose 4 out of them then we whd hv had only one way of doing it , we can prove it by formula also 4c4 = 1 .....) tat is equal to 5 AND we can arrange 4 DIFF balls in 4! ways
as we used AND in above line , e solution wil b like
5 x 4!

c)
all balls same , so ball can b arranged in 1 way only n as all boxes same so we can choose 4 identical boxes out of 5 identical boxes in 5c4=5
so ans is
1x5=5

i dont noe abt formula u mentioned above , our faculty told us to take step by step approach or else this topic can b highly confusing

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 seem to have a little confusion between a) and c).I feel they are the same .In other words, in a) all that is important is to choose 4 boxes from 5.Why is it important , the order in which you put the balls in the boxes as the balls are all similar ?i feel there is no permutation involved....I know this is the wrong answer but im not able to convince myself...so can somebody help....?

hi vinayak
is the formulae for the following
The number of positive integral solution(natural number) for
x1+x2+x3+.........xr=n is (n-1) C (r-1) or (n-r+1)C(r-1)

Nine students are split into three equal teams to develop reports on one of three problems:

shortage of skilled labor, violence in schools, and low standardized test scores. How many

different teams of students are possible?

(A) 5040

(B) 1680

(C) 1512

(D) 504

(E) 168

Nine students are split into three equal teams to develop reports on one of three problems:

shortage of skilled labor, violence in schools, and low standardized test scores. How many

different teams of students are possible?

(A) 5040

(B) 1680

(C) 1512

(D) 504

(E) 168

we can choose 3 students for the first team in C(9,3) ways.
out of the remaining 6 we can choose 3 in C(6,3) ways.
the remiaining can go into the third team.

so the number of ways of choosing 3 teams of 3 each from 9 students is C(9,3)*C(6,3)=1680.

we can choose a problem out of the three in 3 ways.
so the number of ways in which 3 teams can be formed to study one of the three problems is 1680*3=5040.

in case each team had to study a different problem the answe would have been 1680*3!=10080.

bye..

Hey pendyal,

Don't you think a more appropriate answer is the latter one you gave i.e. 1680*3! ( though it is not in the options)?

Also, i think we can even stop at 1680 because the question says "How many teams can be formed?".

But 1680*3 doesn't seem to be too convincing.

1680*3 was only a way to come up with one of the options as the answer.  if i were not given the options i wud have given 1680*3! as the answer. 😃

There are 5 Rock songs, 6 Carnatic songs and 3 Indi pop songs. How many different albums can be formed using the above repertoire if the albums should contain at least 1 Rock song and 1 Carnatic song?

Solve with proper explanation...

There are 5 Rock songs, 6 Carnatic songs and 3 Indi pop songs. How many different albums can be formed using the above repertoire if the albums should contain at least 1 Rock song and 1 Carnatic song?

Solve with proper explanation...

first choose a rock song and a carnatic song.this can be done in 5 ways and 6 ways respectively.
so we can choose a rock song and a carnatic song together in 30 ways.

now of the remaining 12 songs,each song can either be in the album or not be in the album.that is there are 2 ways for each of the 12 songs.
so these 12 songs can be chosen in 2*2*2....*2 (i.e. 2^12).

the total no. of ways becomes 30*(2^12).

bye..

A bin of computer disks contain a supply of disks from 4 different
manufacturers. In how many ways can you choose 6 disks from the bin?

hey solve with explanation........

A bin of computer disks contain a supply of disks from 4 different
manufacturers. In how many ways can you choose 6 disks from the bin?

hey solve with explanation........

each disk chosen from the bin can be chosen in 4 ways.

so the 6 disks can be chosenin 4*4*4*4*4*4 ways i.e 4^6 ways.

bye..

hey but OA is 84

A bin of computer disks contain a supply of disks from 4 different
manufacturers. In how many ways can you choose 6 disks from the bin?

hey solve with explanation........

each disk chosen from the bin can be chosen in 4 ways.

so the 6 disks can be chosenin 4*4*4*4*4*4 ways i.e 4^6 ways.

bye..

Pendyal,
I think you have all the permutations, but what we require here, are the combinations.
I am getting the answer using the following approach:

1) 6 disks are from the same manufacturer(M) = 4 cases
2) Cases with 2 Ms :
a) 5 --> M1 & 1 --> M2 = 12 cases
b) 4 --> M1 & 2 --> M2 = 12 cases
c) 3 --> M1 & 3 --> M2 = 6 cases
3) Cases with 3 Ms:
a) 4 --> M1 & 1 --> M2 & 1 --> M3 = 12 cases
b) 3 --> M1 & 2 --> M2 & 1 --> M3 = 24 cases
c) 2 --> M1 & 2 --> M2 & 2 --> M3 = 4 cases
4) Cases with 4 Ms:
a) 3 --> M1 & 1 --> M2 & 1 --> M3 & 1 --> M4 = 4 cases
b) 2 --> M1 & 2 --> M2 & 1 --> M3 & 1 --> M4 = 6 cases

Hence, adding all the above possible cases,
= 12 + 12 + 6 + 12 + 24 + 4 + 6
= 84

As, the solution was provided, it made my life much easier If the above solution seems correct, please reply, so that if required, I can elaborate further.

Cheers Gaurav

Pendyal,
I think you have all the permutations, but what we require here, are the combinations.
I am getting the answer using the following approach:

1) 6 disks are from the same manufacturer(M) = 4 cases
2) Cases with 2 Ms :
a) 5 --> M1 & 1 --> M2 = 12 cases
b) 4 --> M1 & 2 --> M2 = 12 cases
c) 3 --> M1 & 3 --> M2 = 6 cases
3) Cases with 3 Ms:
a) 4 --> M1 & 1 --> M2 & 1 --> M3 = 12 cases
b) 3 --> M1 & 2 --> M2 & 1 --> M3 = 24 cases
c) 2 --> M1 & 2 --> M2 & 2 --> M3 = 4 cases
4) Cases with 4 Ms:
a) 3 --> M1 & 1 --> M2 & 1 --> M3 & 1 --> M4 = 4 cases
b) 2 --> M1 & 2 --> M2 & 1 --> M3 & 1 --> M4 = 6 cases

Hence, adding all the above possible cases,
= 12 + 12 + 6 + 12 + 24 + 4 + 6
= 84

As, the solution was provided, it made my life much easier If the above solution seems correct, please reply, so that if required, I can elaborate further.

Cheers Gaurav

i didnt understand the solution.cud u please elaborate.
(more specifically in the third case of 2 manufacturers how u got 6 ways and in the other 2 cases how u arrived at the possible combinations was all greek to me.)

hey it seems right as the official answer is 84 .... thatz what i had mentioned earlier ....

Pendyal,
I think you have all the permutations, but what we require here, are the combinations.
I am getting the answer using the following approach:

1) 6 disks are from the same manufacturer(M) = 4 cases
2) Cases with 2 Ms :
a) 5 --> M1 & 1 --> M2 = 12 cases
b) 4 --> M1 & 2 --> M2 = 12 cases
c) 3 --> M1 & 3 --> M2 = 6 cases
3) Cases with 3 Ms:
a) 4 --> M1 & 1 --> M2 & 1 --> M3 = 12 cases
b) 3 --> M1 & 2 --> M2 & 1 --> M3 = 24 cases
c) 2 --> M1 & 2 --> M2 & 2 --> M3 = 4 cases
4) Cases with 4 Ms:
a) 3 --> M1 & 1 --> M2 & 1 --> M3 & 1 --> M4 = 4 cases
b) 2 --> M1 & 2 --> M2 & 1 --> M3 & 1 --> M4 = 6 cases

Hence, adding all the above possible cases,
= 12 + 12 + 6 + 12 + 24 + 4 + 6
= 84

As, the solution was provided, it made my life much easier If the above solution seems correct, please reply, so that if required, I can elaborate further.

Cheers Gaurav

i didnt understand the solution.cud u please elaborate.
(more specifically in the third case of 2 manufacturers how u got 6 ways and in the other 2 cases how u arrived at the possible combinations was all greek to me.)

I think its rite..see if my explantaion is ok....

as he says:

1) 6 disks are from the same manufacturer(M) = 4 cases
all are eitehr from M1 or M2 or M3 or M4

M1 M2 M3 M4
6 0 0 0
0 6 0 0
0 0 6 0
0 0 0 6

2) Cases with 2 Ms :

You select two manufaturers and from the two you can get different combinations like (5,1) (4,2) (3,3)

therefore for (5,1)

M1 M2 M3 M4
5 1 0 0
5 0 1 0
and so on...
Formaula is :select 2 out of 4 manufaturers 4C2
now the two can be intercahnged that is (5,1) to (1,5)
therefore multiply by 2
===>4C2*2= 12

same way for (4,2)
M1 M2 M3 M4
4 2 0 0
4 0 2 0
and so on

same way as before
====>4C2*2=12

for (3,3)
select two 4C2....interchanging (3,3) to (3,3) doesn't matter

hence
====>4C2=6

a) 5 --> M1 & 1 --> M2 = 12 cases
b) 4 --> M1 & 2 --> M2 = 12 cases
c) 3 --> M1 & 3 --> M2 = 6 cases

3) Cases with 3 Ms:

a) 4 --> M1 & 1 --> M2 & 1 --> M3 = 12 cases
select 3 = 4C3
(4,1,1)-(1,4,1)-(1,1,4) two are 1,1 so repetitions in 3! so divide by 2!
3!/2! = 3

b) 3 --> M1 & 2 --> M2 & 1 --> M3 = 24 cases
4C3*3! (all are differnt number 3,2,1
====>24
c) 2 --> M1 & 2 --> M2 & 2 --> M3 = 4 cases
select 3 = 4C3
====> 4

4) Cases with 4 Ms:
a) 3 --> M1 & 1 --> M2 & 1 --> M3 & 1 --> M4 = 4 cases
4!/3! =4
====> 4
b) 2 --> M1 & 2 --> M2 & 1 --> M3 & 1 --> M4 = 6 cases
4!/(2!*2!)
====> 6