6 boys and 6 girls sit in a row at random. Find the probability that 1) The six girls sit together 2) The boys and girls sit alternately.a)1/462b)c)d)Correct answer is option 'A'. Can you explain this answer?

Question

Answer

\Total ways of making sit 12 persons 12!
Considering set of 6 girls as a single group. Within the group girls can be arranged in 6! Ways. Now only among 7 persons—> 6 boys + 1 girl group arrangement needs to be done. Thus, number of ways for the same = 7!.
Therefore, probability = Favourable outcomes/ total number of outcomes = (7!) (6!) / (12!) = 1/1322
We can only make sit boys and girls in alternate manner in two possible ways, boys at even and girls at even. But, among boys group and girls group itself, rearrangement can be done in 6! Ways each.
Therefore, probability =(2)(6!)(6!)/ 12! = 1/462

Answer

1) Total ways of sitting =12! Consider set of 6 girls as 1 group. Now total 7 persons have   to sit So total ways = 7! Now within the group of girls they can be arranged in 6! ways. Probabilty = 7!*6!/12!=1/1322) There are two ways of sitting alternatively. Boys at even places and girls at odd places and vice versa. In both case boys and girls can be arranged in 6!*6! ways.Probability = 6!*6!*2/12! =1/462

Answer

1/7 ---- probablity of six girls together(counted as 1) and remaining 6 guys alternate arrangements--- 6(different arrangements) x remaining 11 ( because alternate so only 6 and not 12)...it's probablity is 1/6*11 =1/66
now final answer multiply both 1/7*1/66 you get option A

Answer

\Total ways of making sit 12 persons 12!
Considering set of 6 girls as a single group. Within the group girls can be arranged in 6! Ways. Now only among 7 persons—> 6 boys + 1 girl group arrangement needs to be done. Thus, number of ways for the same = 7!.
Therefore, probability = Favourable outcomes/ total number of outcomes = (7!) (6!) / (12!) = 1/1322
We can only make sit boys and girls in alternate manner in two possible ways, boys at even and girls at even. But, among boys group and girls group itself, rearrangement can be done in 6! Ways each.
Therefore, probability =(2)(6!)(6!)/ 12! = 1/462

Answer

To find the probability of certain events occurring when 6 boys and 6 girls sit in a row at random, we can use the concept of permutations.

1) The probability that the six girls sit together:

First, let's consider the six girls as a single entity. Now, we have 7 entities in total - the group of six girls and the six boys.

The total number of ways these entities can be arranged is 7!. However, within the group of six girls, they can be arranged among themselves in 6! ways.

Therefore, the total number of arrangements where the six girls sit together is 7! × 6!.

Now, let's consider the total number of arrangements without any restrictions. The total number of ways the 12 individuals can be arranged is 12!.

Therefore, the probability of the six girls sitting together is given by:

P(six girls together) = (7! × 6!) / 12!

Simplifying this expression, we have:

P(six girls together) = (7! × 6!) / 12! = 1/462

Therefore, the correct answer is option A) 1/462.

2) The probability that the boys and girls sit alternately:

In this case, we can consider the boys and girls as two separate groups. The total number of ways these groups can be arranged is 2!.

Within each group, the boys can be arranged among themselves in 6! ways, and the girls can be arranged among themselves in 6! ways.

Therefore, the total number of arrangements where the boys and girls sit alternately is 2! × 6! × 6!.

The total number of arrangements without any restrictions is still 12!.

Therefore, the probability of the boys and girls sitting alternately is given by:

P(boys and girls alternate) = (2! × 6! × 6!) / 12!

Simplifying this expression, we have:

P(boys and girls alternate) = (2! × 6! × 6!) / 12! = 1/132

Therefore, the correct answer for this part is option B) 1/132.

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