Need clarification on permutation/combination problem

[tushar4036]

1.There are 2 brothers among a group of 20 persons.In how many ways can the group be arranged around a circle so that there is exactly one person between the two brothers.

2.How many five digit numbers can be formed using the digits 0,1,2,3,4 and 5 which are divisible by3,without repeating the digit?

[ace_gre]

Q2.

To be divisible by 3, sum of digits should be divisible by 3.
No. of ways of choosing 5 numbers from 6: 6C5 = 6C1= 6

Combinations are as follows:

1) 0,1,2,3,4; 0+1+2+3+4 =10
2) 1,2,3,4,5; 1+2+3+4+5 =15
3) 2,3,4,5,0; 2+3+4+5+0 =14
4) 3,4,5,0,1; 3+4+5+0+1 =13
5) 4,5,0,1,2; 4+5+0+1+2 =12
6) 5,0,1,2,3; 5+0+1+2+3 =11

Only number combinations of #2 and #5 are divisible by 3.

1,2,3,4,5
---------
Permutations possible: 5! = 120

4,5,0,1,2
---------

Permutations possible: 4 x 4 x 3 x 2 x 1 = 96 (since 0 cannot be the first digit)

Total 5 digit numbers divisible by 3 = 120+96 = 216.

Thanks!

[ace_gre]

Q1.

I am not good at circular combinations, but wanted to give this a try..So here is my approach. All, please let me know if you see any flaws!!!

Let the two brothers be B1 and B2.

In a circular arrangement if B1 occupies seat 1, then B2 can sit in seat3 or seat 19 .So for every 20 seats, B1 occupies B2 can sit in 2 seats. So there are 20 x 2= 40.

18 other people can occupy the seat in-between B1 and B2. So total number of seating arrangements = 40 x 18 = 720. ??

is this the correct answer??

thanks!

[nicrey]

Originally Posted by tushar4036

1.There are 2 brothers among a group of 20 persons.In how many ways can the group be arranged around a circle so that there is exactly one person between the two brothers.

2.How many five digit numbers can be formed using the digits 0,1,2,3,4 and 5 which are divisible by3,without repeating the digit?

let say that 3 people (2 brothers and one person) form a group. There are 2 x 18 x 1, of selecting this group. Now we have 16 persons and 1 person (which represents 3 perrsons with 36 ways of selecting. So we have 36 x (16-1)!

(16-1)! circular permutation of 17 persons.

[ace_gre]

nicrey,

can you please explain more clearly.I cannot understand the explanation you gave earlier.

thanks

[feminista]

I understand your logic for Q1 (and that there are 6 possible sets of numbers in the first step), but what operation/relationship are you denoting when you say "6C5=6C1=6"?

Thanks!

[ace_gre]

Try expanding the formula nCr= n!/((n-r)!*r!).
Since the denominator has both r! and n-r!, 6C1=6C5, 6C2=6C4,6C6=6C0 and holds true for all values of n and r.

Hope this helps!

[Curly213]

Quote:
Originally Posted by tushar4036 (Need clarification on permutation/combination problem)
1.There are 2 brothers among a group of 20 persons.In how many ways can the group be arranged around a circle so that there is exactly one person between the two brothers.

2.How many five digit numbers can be formed using the digits 0,1,2,3,4 and 5 which are divisible by3,without repeating the digit?

let say that 3 people (2 brothers and one person) form a group. There are 2 x 18 x 1, of selecting this group. Now we have 16 persons and 1 person (which represents 3 perrsons with 36 ways of selecting. So we have 36 x (16-1)!

(16-1)! circular permutation of 17 persons.


thats a nice calculation but i have some doubts about (16-1)!.... aint it (18-1)! , because we already took 3 persons to get the C number and now we have 17 left ..20-3=17 so i think the end solution should be 36*17!

[Curly213]

Could s.o . confirm if iam right or iam just havin somethin wrong in my mind????

[totymody]

1-
'n' objects can be arranged around a circle in (n - 1)!.
Let there be exactly one person between the two brothers as stated in the question.

If we consider the two brothers and the person in between the brothers as a block, then there will 17 others and this block of three people to be arranged around a circle.

The number of ways of arranging 18 objects around a circle is in 17! ways.

Now the brothers can be arranged on either side of the person who is in between the brothers in 2! ways.

Therefore, the total number of ways 17! * 2 = 2 * 17!.