3 Couples in a Row

Very useful information for 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 !

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:

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

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.

1. 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 $P_1 = 6$ .
2. 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 $P = \dfrac {P_1+P_2}{N} = \dfrac {6+24}{90} = \dfrac {30}{90} = \boxed{\frac{1}{3}}$

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

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

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.