## probability .

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- Md.Nazrul Islam
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Probability that neither of the couples sits together in adjacent chairs = 1 - probability that at least one couple sits togetherMd.Nazrul Islam wrote:one single person and two couple are to be seated at random in a of raw five chairs ,what is the probability that neither of the couples sits together in adjacent chairs .

Now, when at least one couple sits together is possible when exactly 1 couple sit together and exactly 2 couples sit together.

Let us assume that A, B, C, D, E are 5 people, where 1st couple is A-B, 2nd couple is C-D, and single person is E.

**When exactly 1 couple sit together:**A-B, C, D, E

No. of possible ways = 4! * 2! = 4 * 3 * 2 * 2 = 48, but this also include the no. of ways when 2 couples sit together.

So, no. of ways in which A-B can be seated together = 48 - 4! = 48 - (4 * 3 * 2) = 48 - 24 =

**24 ways**

Similarly, no. of ways in which C-D can be seated together =

**24 ways**

**When exactly 2 couples sit together:**A-B, C-D, E

No. of possible ways = 3! * 2! * 2! = 3 * 2 * 2 * 2 =

**24 ways**

So, no. of ways in which at least one couple sits together = 24 + 24 + 24 = 72

Since there are a total of 5 persons, so they can be seated in 5! = 5 * 4 * 3 * 2 = 120 ways

Probability that at least one couple sits together = 72/120 = 3/5

Therefore, probability that neither of the couples sits together in adjacent chairs = 1 - 3/5 = [spoiler]

**2/5**[/spoiler]

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Let's say that we have couple AB, couple CD, and lonely person E.Md.Nazrul Islam wrote:one single person and two couple are to be seated at random in a of raw five chairs ,what is the probability that neither of the couples sits together in adjacent chairs .

**Total arrangements = (arrangements with AB together) + (arrangements with CD together) - (arrangements with both AB and CD together) + (arrangements with neither AB nor CD together)**

The big idea is to SUBTRACT THE OVERLAP.

When we count the arrangements in which AB sit together and those in which CD sit together, the arrangements in which

*both*AB and CD sit together -- the OVERLAP -- gets counted twice.

Thus, we need to SUBTRACT THE OVERLAP -- the arrangements in which both AB and CD sit together -- so that these arrangements are not double-counted.

Total arrangements = 5! = 120

AB together:

Number of ways to arrange the 4 elements AB, C, D, and E = 4! = 24.

Since AB can be reversed to BA, we multiply by 2:

2*24 = 48.

CD together:

Number of ways to arrange the 4 elements CD, A, B and E = 4! = 24.

Since CD can be reversed to DC, we multiply by 2:

2*24 = 48.

Both AB and CD together:

Number of ways to arrange the 3 elements AB, CD, and E = 3! = 6.

Since AB can be reversed, CD can be reversed, and both AB and CD can be reversed, we multiply by 4:

4*6 = 24.

Plugging these values into the equation above, we get:

120 = 48 + 48 - 24 + N

120 = 72 + N

N = 48

Thus, P(neither couple sit together) = 48/120 = 2/5.

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**Concept Problem**: Now, let's say 4 more guests show up, another couple and two loners. So total there are 9 chairs for 9 guests - 3 couples and 3 stags. If the guests are seated randomly, what is the new probability that none of the couples will sit together?

Guest list: AB, CD, EF, G, H, I

**Total arrangements = (arr. AB together) + (arr. CD together) + (arr. EF together) - (arr. AB + CD together) - (arr. AB + EF together) - (arr. CD + EF together) - (arr. AB + CD + EF together) + (arr. none are together) = 9! = 362, 880**

**AB together**: We have 8 elements - AB, C, D, E, F, G, H, I - so total arrangements = 8! = 40, 320. Multiply by 2 to account for AB->BA switch = 80,640.

**CD together**: same reasoning = 80,640.

**EF together**: same reasoning = 80,640.

**ABCD together**: We have 6 elements - ABCD, E, F, G, H, I - so total arrangements = 6! = 720. Multiply by 2 to account for AB->BA switch, and by another 2 for the CD->DC switch = 720*4 = 2,880.

**ABEF together**: same reasoning = 2,880.

**CDEF together**: same reasoning = 2,880.

**ABCDEF together**: We have 4 elements - ABCDEF, G, H, I - so total arrangements = 4! = 24. Multiply by 2 for each couple's switch = 24 * 8 = 192.

Return to original: Total arrangements = (arr. AB together) + (arr. CD together) + (arr. EF together) - (arr. AB + CD together) - (arr. AB + EF together) - (arr. CD + EF together) - (arr. AB + CD + EF together) + (arr. none are together) = 80,640 + 80,640 + 80,640 - 2,880 - 2,880 - 2,880 - 192 + (arr. none are together) = 362, 880

Arrangements none together = 129,792

**Probability none sit together = 129,792/362,880 = .357**

Don't worry about running the actual numbers, I just want to check the conceptual framework. Thanks!