What is the number of ways that $5$ boys and $5$ girls can be seated in a row so that boys and girls sit alternately?
$28800$
Understanding the Alternating Seating Arrangement
The problem asks for the number of ways to arrange 5 boys and 5 girls in a single row such that no two boys sit together and no two girls sit together. This means they must alternate.
Analyzing the Two Possible Alternating Patterns
For boys and girls to sit alternately in a row of 10 seats (5 boys + 5 girls), there are only two possible patterns for the arrangement:
- Pattern 1: The row starts with a boy (B), followed by a girl (G), and so on. This looks like: B G B G B G B G B G
- Pattern 2: The row starts with a girl (G), followed by a boy (B), and so on. This looks like: G B G B G B G B G B
Calculating Arrangements for Each Pattern
We need to calculate the number of ways to arrange the boys and girls within each pattern. This involves permutations, as the order in which individuals are seated matters.
Case 1: Arrangement Starts with a Boy (B G B G B G B G B G)
In this pattern, the 5 positions designated for boys (1st, 3rd, 5th, 7th, 9th) can be filled by the 5 boys in $5!$ ways. Similarly, the 5 positions designated for girls (2nd, 4th, 6th, 8th, 10th) can be filled by the 5 girls in $5!$ ways. The total number of arrangements for this specific pattern is the product of the ways to arrange boys and the ways to arrange girls. Ways for Case 1 = (Ways to arrange 5 boys) $\times$ (Ways to arrange 5 girls) = $5! \times 5!$.
Case 2: Arrangement Starts with a Girl (G B G B G B G B G B)
In this pattern, the 5 positions designated for girls (1st, 3rd, 5th, 7th, 9th) can be filled by the 5 girls in $5!$ ways. The 5 positions designated for boys (2nd, 4th, 6th, 8th, 10th) can be filled by the 5 boys in $5!$ ways. The total number of arrangements for this specific pattern is also the product of the ways to arrange girls and the ways to arrange boys. Ways for Case 2 = (Ways to arrange 5 girls) $\times$ (Ways to arrange 5 boys) = $5! \times 5!$.
Total Number of Ways for Alternating Seating
Since Case 1 and Case 2 are the only two possible ways for the boys and girls to sit alternately, the total number of ways is the sum of the ways calculated for each case.
Total Ways = Ways (Case 1) + Ways (Case 2)
Total Ways = ($5! \times 5!$) + ($5! \times 5!$)
Total Ways = $2 \times (5! \times 5!)$
Step-by-Step Calculation
Let's calculate the final value:
- First, calculate the value of $5!$ (5 factorial):
$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$
- Next, calculate $5! \times 5!$:
$5! \times 5! = 120 \times 120 = 14400$
- Finally, multiply by 2 to account for both possible starting patterns (starting with a boy or starting with a girl):
Total Ways = $2 \times 14400 = 28800$
Conclusion on Seating Arrangements
Therefore, there are $28800$ distinct ways to seat 5 boys and 5 girls in a row such that they sit alternately.
Important Questions from Permutations and Combinations
The number of ways of choosing 21 objects out of 42 objects of which 21 are identical and the remaining 21 are distinct, is:
If nPr = 720 and nCr = 120, then the value of r is:
For a social work, 7 men and 6 women gave their nominations. The committee is formed to select 5 people from the nominated persons in such a way that atleast 3 men are there in the final team. Find the number of ways in which the people can be selected.
The largest coefficient of ( x + 1)20 is:
If 2n+1Pn–1: 2n–1Pn = 3 : 5, then what is the value of n?
