Circular arrangements (1 Viewer)

=)(= · Oct 3, 2021

Pls help with q6 d

ExtremelyBoredUser · Oct 3, 2021

At least 3 odd numbers are together: All 4 odd numbers together + 3 odd numbers together. Select any 3 odd numbers in 4C3 ways and arrange the group in 3! ways, the 4th odd number can be arranged in 3 ways and the rest 4 even in 4! Ways. So, total ways 576+4×3!×3×4!=576+1728=2304

Use cases but you have to be weary of conditions of the cases:

At least 3 odd number can be broken up to 2 cases -> All 4 possible odds together + only 3 odd numbers together.

1st Case:
You can choose 3 odd numbers in 4C3 ways (Max 4 odd numbers, only wanting 3) and you can arrange them inside their group in 3! ways, whereas the last odd number can be arranged only in 3 ways so it does not clump together with the 3 odd number group and the rest of the numbers can be arranged in 4! ways.

4C3 * 4! * 3! = 576 ways for first case

2nd Case:
Unlike the 1st case you have to ensure that the fixed spot, so the top of the circle, is not occupied by an odd number since they are all clumped together. Fix the spot on the top with an even number and separate the even and odd numbers in distinct groups.

4 spots for the 4 odd numbers and you can arrange them in 4! ways.
3 spots for 3 even numbers and you can arrange them in 3! ways.

3 * 3! * 4 * 4! = 1728 ways

Therefore total cases is 576 + 1728 = 2304 ways.

Edit:
Cases are the other way around. Check post below.

CM_Tutor · Oct 3, 2021

@ExtremelyBoredUser, are your cases around the wrong way?

Case 1: 3 odd together, one separate:

$\text{Choose the 3 odds from 4 available in}\ \binom{4}{3} = 4\ \text{ways}$
$\text{This block of 3 odd numbers can be ordered in}\ 3!\ \text{ways}$
$\text{We now have 6 objects to place in a circle - one "block" of three odd numbers, a fourth odd number, and four different even numbers}$
$\text{Place this block into the circle in}\ 1\ \text{way, as all positions on a circle are identical for the initial placement}$
$\text{Place the remaining odd number in}\ 3\ \text{ways, as two of the five remaining positions are forbidden as making a four-odd-nummbers-together block}$
$\text{Place the four even numbers into the four remaining positions in}\ 4!\ \text{ways}$

$\text{So, total arrangements under Case 1}\ = 4 \times 3! \times 1 \times 3 \times 4! = 1,728\ \text{ways}$

Case 2: All four odd numbers together:

$\text{We have a block of four odd numbers, which can be selected in only 1 way, and can be ordered in}\ 4!\ \text{ways}$
$\text{We now have 5 objects to place in a circle - one "block" of four odd numbers and four different even numbers}$
$\text{Place this block into the circle in}\ 1\ \text{way, as all positions on a circle are identical for the initial placement}$
$\text{Place the four even numbers into the four remaining positions in}\ 4!\ \text{ways}$

$\text{So, total arrangements under Case 2}\ = 4! \times 1 \times 4! = 576\ \text{ways}$

$\text{So, total arrangements under both cases}\ = 1,728 + 576 = 2,304\ \text{ways}$

CM_Tutor · Oct 3, 2021

This could also be done by placing the block of 3, and then arranging the other five, to get

$\binom{4}{3} \times 3! \times 1 \times 5! = 2,880\ \text{ways}$

so long as it is recognised that some the cases like odd block is 3, 5, 7, and then 1 placed before it duplicates odd block 1, 3, 5, with 7 placed after it. So, there has been a duplication of

$\binom{4}{3} \times 3! \times 1 \times 1 \times 4! = 576\ \text{ways}$

$\text{Thus, total number of different arrangements}\ = 2,880 - 576 = 2,304\ \text{ways}$

ExtremelyBoredUser · Oct 3, 2021

CM_Tutor said:
@ExtremelyBoredUser, are your cases around the wrong way?

Case 1: 3 odd together, one separate:

$\text{Choose the 3 odds from 4 available in}\ \binom{4}{3} = 4\ \text{ways}$

$\text{This block of 3 odd numbers can be ordered in}\ 3!\ \text{ways}$

$\text{We now have 6 objects to place in a circle - one "block" of three odd numbers, a fourth odd number, and four different even numbers}$

$\text{Place this block into the circle in}\ 1\ \text{way, as all positions on a circle are identical for the initial placement}$

$\text{Place the remaining odd number in}\ 3\ \text{ways, as two of the five remaining positions are forbidden as making a four-odd-nummbers-together block}$

$\text{Place the four even numbers into the four remaining positions in}\ 4!\ \text{ways}$

$\text{So, total arrangements under Case 1}\ = 4 \times 3! \times 1 \times 3 \times 4! = 1,728\ \text{ways}$

Case 2: All four odd numbers together:

$\text{We have a block of four odd numbers, which can be selected in only 1 way, and can be ordered in}\ 4!\ \text{ways}$

$\text{We now have 5 objects to place in a circle - one "block" of four odd numbers and four different even numbers}$

$\text{Place this block into the circle in}\ 1\ \text{way, as all positions on a circle are identical for the initial placement}$

$\text{Place the four even numbers into the four remaining positions in}\ 4!\ \text{ways}$

$\text{So, total arrangements under Case 2}\ = 4! \times 1 \times 4! = 576\ \text{ways}$

$\text{So, total arrangements under both cases}\ = 1,728 + 576 = 2,304\ \text{ways}$

Yeah, you're right rechecking, I checked my working out for the question and it seems I interpreted wrongly/illogically. Thanks for pointing that out

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