Six boys and six girls sit in a row...

MathematicsClassical Definition of ProbabilityJEE Advanced 1979Moderate

Six boys and six girls sit in a row randomly. Find the probability that (i) the six girls sit together (ii) the boys and girls sit alternately.

Visualized Solution (Hindi)

Defining the Sample Space

Total number of people: $6 boys + 6 girls = 12$
Total number of random arrangements (Sample Space): $n (S) = 12!$

Part (i): The String Method

Condition: All 6 girls sit together.
Treat the 6 girls as one single unit (bundle).
Remaining items: 6 boys + 1 girl-bundle = 7 units.

Arranging the Bundle

Ways to arrange 7 units: $7!$
Ways to arrange 6 girls within their bundle: $6!$
Total favorable arrangements: $n (E_{1}) = 7! \times 6!$

Calculating Probability for Part (i)

$P (Girls together) = \frac{7 ! \times 6 !}{12 !}$
$= \frac{7 ! \times 720}{12 \times 11 \times 10 \times 9 \times 8 \times 7 !}$
$= \frac{720}{95040} = \frac{1}{132}$

Part (ii): Alternating Arrangement

Condition: Boys and girls sit alternately.
Two possible patterns:
1. $B GB GB GB GB GB G$
2. $GB GB GB GB GB GB$

Calculating Ways for Pattern 1

Pattern 1: $B$ at positions 1, 3, 5, 7, 9, 11 and $G$ at 2, 4, 6, 8, 10, 12.
Ways for boys: $6!$
Ways for girls: $6!$
Total for Pattern 1: $6! \times 6!$

Total Favorable Ways for Part (ii)

Pattern 2: $G$ at odd positions, $B$ at even positions.
Total for Pattern 2: $6! \times 6!$
Total favorable arrangements: $n (E_{2}) = 2 \times (6! \times 6!)$

Calculating Probability for Part (ii)

$P (Alternate) = \frac{2 \times 6 ! \times 6 !}{12 !}$
$= \frac{2 \times 720 \times 6 !}{12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 !}$
$= \frac{1440}{665280} = \frac{1}{462}$

Summary and Key Takeaway

Key Takeaways:
Use the String Method for 'together' conditions.
Consider all possible starting positions for alternating arrangements.
Probability (i) = $1/132$
Probability (ii) = $1/462$

00:00 / 00:00

Conceptually Similar Problems

MathematicsClassical Definition of ProbabilityJEE Advanced 1996Moderate

View in:English Hindi

In how many ways three girls and nine boys can be seated in two vans, each having numbered seats, 3 in the front and 4 at the back? How many seating arrangements are possible if 3 girls should sit together in a back row on adjacent seats? Now, if all the seating arrangements are equally likely, what is the probability of 3 girls sitting together in a back row on adjacent seats?

MathematicsClassical Definition of ProbabilityJEE Advanced 2014Easy

View in:English Hindi

Three boys and two girls stand in a queue. The probability, that the number of boys ahead of every girl is at least one more than the number of girls ahead of her, is

(A)

1/2

(B)

1/3

(C)

2/3

(D)

3/4

Answer:A

MathematicsLinear PermutationsJEE Advanced 1983Easy

View in:English Hindi

$m$ men and $n$ women are to be seated in a row so that no two women sit together. If $m > n$ , then show that the number of ways in which they can be seated is $\frac{m ! ( m + 1 )!}{( m - n + 1 )!}$

MathematicsConditional ProbabilityJEE Advanced 2007Moderate

View in:English Hindi

One Indian and four American men and their wives are to be seated randomly around a circular table. Then the conditional probability that the Indian man is seated adjacent to his wife given that each American man is seated adjacent to his wife is

(A)

1/2

(B)

1/3

(C)

2/5

(D)

1/5

Answer:C

MathematicsCircular PermutationsJEE Main 2003Easy

View in:English Hindi

The number of ways in which 6 men and 5 women can dine at a round table if no two women are to sit together is given by

(A)

7! \times 5!

(B)

6! \times 5!

(C)

30!

(D)

5! \times 4!

Answer:B

MathematicsLinear PermutationsJEE Advanced 1982Easy

View in:English Hindi

Eight chairs are numbered 1 to 8. Two women and three men wish to occupy one chair each. First the women choose the chairs from amongst the chairs marked 1 to 4 ; and then the men select the chairs from amongst the remaining. The number of possible arrangements is

(A)

^{6} C_{3} \times^{4} C_{2}

(B)

^{4} P_{2} \times^{6} P_{3}

(C)

^{4} C_{2} +^{4} P_{3}

(D)

none of these

Answer:B

MathematicsLinear PermutationsJEE Advanced 2015Moderate

View in:English Hindi

Let $n$ be the number of ways in which 5 boys and 5 girls can stand in a queue in such a way that all the girls stand consecutively in the queue. Let $m$ be the number of ways in which 5 boys and 5 girls can stand in a queue in such a way that exactly four girls stand consecutively in the queue. Then the value of $\frac{m}{n}$ is

Answer:5

MathematicsCombinations and SelectionJEE Advanced 1991Easy

View in:English Hindi

Eighteen guests have to be seated, half on each side of a long table. Four particular guests desire to sit on one particular side and three others on the other side. Determine the number of ways in which the sitting arrangements can be made.

Answer:

^{11} C_{5} \times 9! \times 9!

MathematicsClassical Definition of ProbabilityJEE Advanced 1978Easy

View in:English Hindi

Balls are drawn one-by-one without replacement from a box containing 2 black, 4 white and 3 red balls till all the balls are drawn. Find the probability that the balls drawn are in the order 2 black, 4 white and 3 red.

Answer:

\frac{1}{1260}

MathematicsCombinations and SelectionJEE Advanced 2016Easy

View in:English Hindi

A debate club consists of 6 girls and 4 boys. A team of 4 members is to be selected from this club including the selection of a captain (from among these 4 members) for the team. If the team has to include at most one boy, then the number of ways of selecting the team is

(A)

380

(B)

320

(C)

260

(D)

Answer:A