Permutations and Combinations

[pendyal]

Originally Posted by ritika_j

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..

[ritika_j]

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........

[pendyal]

Originally Posted by ritika_j

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..

[ritika_j]

hey but OA is 84

[GauravShah]

Originally Posted by pendyal

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]

Originally Posted by GauravShah

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.)

[ritika_j]

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

[meena8204]

Originally Posted by pendyal

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

so add up .....u get the answer

-meena

[pendyal]

thanx meena for that clarification.

bye...

[ritika_j]

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?

i think this can be done by using this formula (n+r-1)C(r-1)=(6+4-1)c(4-1)
=9C3=84