Hi,
Can anyone suggest me?
In how many ways we can arrange three digits say 1,2,3 in a row of six so that no two same digits are together?
Any help is appreciated.
Thanks!!
Ashish
Construct a diagram with your row of six as underscores
___ X ___ X ___ X ___ X ___ X ___
The first underscore can have any of the three integers
3 X ____ X ____ X ____ X ____ X ____
The second underscore can have any of the remaining two integers
3 X 2 X ____ X ____ X ____ X ____
The remaining underscore can be filled with the integer 2, since the possibilities are the integers that are not the one integer to the left of it.
so 3 X 2 X 2 X 2 X 2 X 2 = 3 * 2^5 = 96
You have three choices for the first digit. For the next digit, there are only two choices, because the second digit must be different than the first. Similarly for the third, and the fourth, and so on. Thus, the total number of arrangements is
$\displaystyle 3\cdot2^5=96.$
Thanks for the quick reply.
Actually I am trying to solve the below question?
Three variant of GMAT paper are to be given to 12 students(so that each variant is used for 4 students) and In how many ways can the student be placed in the two rows of six each so that there should be no identical variant side by side and the student sitting one behind the other should have the same variant?
kindly help me out in this?
Thanks
There is little modification in the question mentioned in the first post
sorry I missed that earlier.
In how many ways we can arrange 1,2,3,1,2,3 in a row of six so that no two same digits are together?
I think we can do in the manner mentioned below.
We can select the first digit in the 3 ways (from 1,2,3)
This can be done in 3 ways as mentioned above.let's say it is 1
1 _ _ _ _ _
Now we have to select the next digit from 2,3 as per the condition.
this can be done in 2 ways.let's say it is 2.
1 2 _ _ _ _
we have to select the third digit either from 1 or 3 that is again selected in the 2 ways.
let's say third digit is 1.
now our number becomes.
1 2 1 _ _ _
For 4th digit we have the choice of either 2 or 3 but we cannot select 2 as if we select 2 then two consecutive 3 are placed in last two position which is contrary to what is required.
so the 4 digit is selected in 1 way
1 2 1 3 _ _
similarly the fifth and the last position.
Hence the total number of ways of selecting the numbers are
3 *2*2*1*1*1 = 12 ways
Please let me know if the way I am thinking is right or not.
suggestions are welcome.
Thanks,
Ashish
I like using this calculator for working out permutations and calculations.
Combinations and Permutations Calculator
Hello a69356Call the three variants A, B and C. Work out first the number of ways in which these three variants can be placed in the twelve seats, and then, for each selection, arrange each group of 4 students within their 4 seats.
We note that row 2 will be identical to row 1 as far as the position of the variant papers is concerned, since equal variants must be placed one behind the other.
So we may choose 2 seats from the 6 for the variant A papers in $\displaystyle \binom{6}{2}$ = 15 ways. Of these, $\displaystyle \binom{5}{1}$ = 5 will have two seats side-by-side. So the places for variant A may be chosen in 10 ways. In 5 of these, there will be 3 or 4 adjacent seats left vacant - see below, where these are marked with (*):
A . A . . . (*)
A . . A . .
A . . . A . (*)
A . . . . A (*)
. A . A . .
. A . . A .
. A . . . A (*)
. . A . A .
. . A . . A
. . . A . A (*)
In each of the five marked (*), there are 2 ways of placing papers B and C so that no two of the same variant are adjacent. In the remaining 5, there are 4 ways of placing B & C. So the total number of ways of placing all three sets of papers is 5 x 2 + 5 x 4 = 30.
Having chosen which seats are to be occupied by each variant, there are then 4! ways of arranging the 4 variant A students in their seats; 4! for the B's and 4! for the C's.
So the total number of ways of seating the students altogether is 30 x 4! x 4! x 4! = 414720.
Grandad
Hello PlatoClearly, there are two posts that have been merged here. The OP amplified his first question (in which he talked about arranging 1, 1, 2, 2, 3, 3 in a line with no two identical digits adjacent to one another) into the question about GMAT, in which he introduced the wording that you queried:
So, to answer your question above: No, on two counts:
- Because you cannot have two identical variants adjacent to one another in the same row. Even if you replace 'should' by 'could' this still rules out AAAABB.
- Because I interpret the word 'should' to mean 'must'. 'This student should sit here' means, to me, 'This student is obliged to sit here'. So in row 2, a student with a variant A paper should (i.e. is obliged to) sit behind another student with an A paper. This interpretation is, I think, confirmed by the OP's initial problem (which I have only just seen) in which he refers to just six items in one row.
(The idea of these seating requirements is presumably to reduce the possibility of students cheating.)
Since there are only 30 arrangements, it is relatively simple to list them. I do so here, using the method outlined in my first posting, where I place the A's first and then arrange the B's and C's in the remaining places. I have adopted the strategy of placing an A in the first available space on the left, and then working from left to right to find all possible positions of the second A. Then, within each of these 'A' selections, I have placed the B's again working from left to right to find all possible positions.
The ordered pairs above each group indicate the positions occupied by the A's, and * indicates A's in a pattern that leaves only two possibilities for B & C, the others yielding four possible positions for B & C (as per my first posting).
(1, 3)
ABACBC *
ACABCB *
(1, 4)
ABCABC
ABCACB
ACBABC
ACBACB
(1, 5)
ABCBAC *
ACBCAB *
(1, 6)
ABCBCA *
ACBCBA *
(2, 4)
BACABC
BACACB
CABABC
CABACB
(2, 5)
BABCAC
BACBAC
CABCAB
CACBAB
(2, 6)
BACBCA *
CABCBA *
(3, 5)
BCABAC
BCACAB
CBABAC
CBACAB
(3, 6)
BCABCA
BCACBA
CBABCA
CBACBA
(4, 6)
BCBACA *
CBCABA *
Grandad
Well I think that I now see the confusion.
In the schools of US, I think it is safe to say that:
ABACBC
ACABCB
would be considered six rows of two each whereas
A B
A B
A C
A C
B C
B C
would be considered two rows of six each.
I assumed he was dealing with the test given for graduate business school admission in this country.
What is the GMAT?
Here is what I know as the GMAT: Prep smarter, score higher—guaranteed, or your money back!.
I would like to see the actual question.