The answer is 0.3333.

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All three spouses: 3! x 2^3 = 48 is easier to think. Did you get 240 only from 720 - (48 + 144 + 288) ? Can you elaborate for how exactly two spouses and exactly one spouse are calculated despite explanations by others which seem complicated?

Lu Chee Ket
- 5 years, 7 months ago

Number of ways spouses can be seated = 6! Let anyone of 6 occupy first chair. Then second one can be occupied by any of the remaining 4. Now there are two possibilities. Firstly, spouse of the first one is seated on third chair. Then then there are only two ways for the remaining one spouse and one couple to sit as spouse has to be sandwiched between the couple. Secondly, we consider that the first chair occupant's spouse does not occupy third chair. Obviously second person's spouse is also not allowed. So we have only 2 choices from the third couple and then rest of the three chairs have no couple. So we have to leave aside third chair occupant's spouse that is two choices for each of fourth and fifith chairs. So finally the favourable cases are: 6 x 4 (2 + 2 x 2 x 2) = 240. As 240 when divided by 6! gives 1/3, the given probability is 0.333

Rajen Kapur
- 6 years, 8 months ago

Let M be a man and W be a woman. If women are separated, then, there is at least a man between each woman so, at least:

WMWMW

There's a man left to be seated, and he can choose among 6 places. (wmMwmw for example)

As there are 3! possible permutations for the women and 3! possible permutations for the men, the solution is:

$\frac{6*3!*3!} {6!} = \frac {1}{3}$

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Let the three couples be $AA$ , $BB$ and $CC$ .

The total of ways of seating $N = \frac {6!} {2!^3} = \frac {720} {8} = 90$ .

There are two main ways of seating so that no couple is sitting together.

- The first three seats have a couple sitting at first and third seats. Then the couple in last three seats must sit in fourth and last seats. For example: $"ABACBC"$ . It should be noted that for each seating of first three seats, there is only one way of seating. And there are only $3! = 6$ ways of seating the first seats. Therefore, the total number of ways of seating for this case $P_1 = 6$ .
- The first three seats are taken by one from each couple $ABC$ . In this case, for each way of seating the first three seats, there are four ways of seating the last three seats. And there are $3!=6$ ways of seating the first three seats. Therefore, the total of ways of seating the first three seats for this case $P_2 = 6 \times 4 = 24$ .

Therefore, the probability that none of the couples are seated together $P = \dfrac {P_1+P_2}{N} = \dfrac {6+24}{90} = \dfrac {30}{90} = \boxed{\frac{1}{3}}$

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Typo! I think you wanna say $\frac{\color{#D61F06}{6!}}{2!^{3}}$ in starting line.

Pranjal Jain
- 6 years, 6 months ago

3 couple seating arrangement in a straight line

Couple A has A1 & A2 as the spouses Couple B has B1 & B2 Couple C has C1 & C2

6 people can arrange in 6! Ways = 720

3 couples can arrange themselves in 3! = 6 ways Each couple can change positions, so total ways they can sit is 6×2×2×2 = 48

Possibility of all 3 being together is 48/ 720 =0.0666

-Exactly 2 couple together-

Let's say A & B are together C1 C2 are seperate. Their arrangement can be

A C1 B C2

B C1 A C2

C1 A C2 B

C1 B C2 A

C1 A B C2

C1 B A C2

6 ways. Within each couple 2 arrangements possible. 6×2×2×2 =48

Here we have considered A & B couple as together. A C or BC can be together too. So total ways for exactly 2 couples together is 48 × 3 = 144.

Probability of exactly 2 together is 144/ 720 = 0.2

Exactly one couple together.

A stays together B1 B2 C1 C2 are separate. Possible arrangements

B1 C1 B2 C2 A

B1 C1 B2 A C2

B1 C1 A B2 C2

B1 A C1 B2 C2

B1 C1 A C2 B2

C1 B1 C2 B2 A

C1 B1 A B2C2

A B1 C1 B2C2

C1 B1 C2 A B2

C1 B1 A C2 B2

C1 A B1 C2 B2

A C1 B1 C2 B2

12 arrangements. Within each couple 2 seating arrangements 12 ×2×2×2 = 96

B & C can stay together too so total ways of exactly 2 together 96 × 3 = 288

Exactly 3 together 3!×(2×2×2) = 48

Exactly 2 together 12×(2×2)×3 = 144
Exactly 1 together 40×(2)×3 = 288

At least 1 together 480 None together (720 - 480) = 240

So the possibilities are

None together 0.3333

At least 1 together 0.6666

Exactly 1 together 0.4

Exactly 2 together 0.2

Exactly 3 together 0.0666

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Let the 3 couples be

1,2

3,4

5,6

Now fix 1 in the first seat.

A) 2 can sit in third seat. Put 3 in between in second seat. The others can only sit as 546 or 645 (2x). 1 and 2 can exchange seats (2x). Each of the four 3, 4, 5, 6 can sit in the second seat (4x). Total number of ways = 2x2x4 = 16.

B) 2 can sit in the fourth seat.

i)3,5 in second and third seat and 4, 6 in fifth and sixth seat. They can exchange seats with each other (2x2x). The couples can exchange places (2x). 1 and 2 can exchange places (2x). Total ways = 2x2x2x2 = 16 ways.

ii) 3, 6 in second and third seat and 4,5 in fifth and sixth seat. Same as i) total of 16 ways.

C) 2 can sit in the fifth seat. Same situation as A) Total of 16 ways.

D) 2 can sit in sixth seat.

i)Now 3,5 and 4,6 sit in the middle seats. The couples can exchange places(2x). They can each exchange places but simultaneously 5364. (2x) 1 and 2 can exchange places (2*). Total of 2x2x2= 8 ways.

ii) 3,6 and 4,5 can sit in the middle seats . Same situation as above , total of 8 ways.

Total number of ways fixing 1 in first seat = 16+16+16+16+8+8 = 80 ways.

We can choose one of 3 couples in 3C1 ways = 3 ways.(3x)

Final total = 80x3= 240 ways.

Number of ways 6 people can sit in a row = 6!= 720

Probability of none of the spouses together = 240/720 = 0.3333

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tell me why is my solution is incorrect...

first we will consider in how many ways the spouses sits together which is :

consider all of them as 3 different spouse pairs,

Thus we can arrange them in 3! ways then we can arrange all the spouses in 2! X 2! X 2! ways... thus total ways sums upto 48 ..

Therefore the probability that couples sit together is 48/720...

=0.0666666......

Therefore the probability of two spouses not sitting together is;

1-0.06666666....

=0.933333........

anonymous suomynona
- 6 years, 8 months ago

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The probability you calculated will include the possibilities in which two pairs , one pair and no pair seating together. But the required probability was only when none of the spouses are seating together. Got it !

Sandeep Bhardwaj
- 6 years, 8 months ago

Introduce a more interesting way of looking at this:

PROGRAM Scientific_way; {Success: 240; Failure: 480; Sum: 720}

USES CRT;

VAR a, b, c, d, e, f: 1..6; p, q, r, s, t, u: 1..6; Ps: ARRAY[1..6] OF EXTENDED; count, success, failure: 0..720;

FUNCTION Matched(i, j: EXTENDED): BOOLEAN; BEGIN IF i+j=1.0 THEN Matched:=TRUE ELSE Matched:=FALSE END;

FUNCTION Unmatched(i, j: EXTENDED): BOOLEAN; BEGIN IF i+j=1.0 THEN Unmatched:=FALSE ELSE Unmatched:=TRUE END;

BEGIN CLRSCR; count:=0; success:=0; failure:=0; Ps[1]:=0.0; Ps[2]:=1.0; Ps[3]:=0.5; Ps[4]:=0.5; Ps[5]:=0.2; Ps[6]:=0.8;

```
FOR p:=1 TO 6 DO
BEGIN
a:=p;
FOR q:=1 TO 6 DO
BEGIN
IF q<>a THEN
BEGIN
b:=q;
FOR r:=1 TO 6 DO
BEGIN
IF (r<>a) AND (r<>b) THEN
BEGIN
c:=r;
FOR s:=1 TO 6 DO
BEGIN
IF (s<>a) AND (s<>b) AND (s<>c) THEN
BEGIN
d:=s;
FOR t:=1 TO 6 DO
BEGIN
IF (t<>a) AND (t<>b) AND (t<>c) AND (t<>d) THEN
BEGIN
e:=t;
FOR u:=1 TO 6 DO
BEGIN
IF (u<>a) AND (u<>b) AND (u<>c) AND (u<>d) AND (u<>e) THEN
BEGIN
f:=u;
INC(count);
IF Unmatched(Ps[p],Ps[q]) AND Unmatched(Ps[q],Ps[r]) AND Unmatched(Ps[r],Ps[s]) AND Unmatched(Ps[s],Ps[t])
AND Unmatched(Ps[t],Ps[u]) THEN
INC(success);
IF Matched(Ps[p],Ps[q]) OR Matched(Ps[q],Ps[r]) OR Matched(Ps[r],Ps[s]) OR Matched(Ps[s],Ps[t])
OR Matched(Ps[t],Ps[u]) THEN
INC(failure);
WRITE(a, ' ', b, ' ', c, ' ', d, ' ', e, ' ', f, ' ', success, ' ', failure, ' ', count);
READLN
END
END
END
END
END
END
END
END
END
END
END;
WRITELN;
WRITELN('Success: ', success, '; Failure: ', failure, '; Sum: ', count);
WRITE('Probability of success = ', 1.0*success/count:1:18);
REPEAT UNTIL READKEY=#27
```

END.

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A scientific method allows people who don't like mathematics but science to understand the answering to this question. We can see, judge and then count one by one without thinking about factorial of mathematics. Yet, a very immediate and certain answer can be arrived.

0+1 = 1, 0.5 + 0.5 = 1 and 0.2 + 0.8 = 1 are closer to base 10. Generally, we can make 0.1 + 0.9 = 1, 0.3 + 0.7 = 1 and etc as well for logic like this.

Lu Chee Ket
- 5 years, 7 months ago

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Very useful informationfor all such type of question :3 couples are to be seated in a row of 6 chairs :No. of ways in which all spouses can be seated with

no restrictions$=\boxed{720}$No. of ways in which all three spouses are seating together $=\boxed{48}$

No. of ways in which exactly two spouses are seating together $=\boxed{144}$

No. of ways in which exactly one spouses is seating together $=\boxed{288}$

No. of ways in which none of the spouses is seating together $=\boxed{240}$

Enjoy !