Question

1. in how many ways can 5 adults and 6 children be arranged in a single line

a) without restrictions?
= 86400

b) if children and adults are alternated?

c) if the adults are all together
and the children are all together?

d) if the children are all together?

Explanation:

Part (a): Without Restrictions

Step 1: Total number of people

We have 5 adults and 6 children, so total people \( n = 5 + 6=11 \).

Step 2: Permutations of \( n \) people

The number of permutations of \( n \) distinct objects is \( n! \). So the number of ways to arrange 11 people is \( 11! \).
\( 11! = 11\times10\times9\times8\times7\times6\times5\times4\times3\times2\times1 = 39916800 \)

Step 1: Determine the starting position

Since there are 6 children and 5 adults, we must start with a child (to alternate: C, A, C, A, ..., C).

Step 2: Arrange children

Number of ways to arrange 6 children: \( 6! \).

Step 3: Arrange adults

Number of ways to arrange 5 adults: \( 5! \).

Step 4: Total arrangements

Since we start with a child, the total number of arrangements is \( 6! \times 5! \).
\( 6! = 720 \), \( 5! = 120 \), so \( 6! \times 5! = 720\times120 = 86400 \)

Step 1: Treat groups as single entities

Treat the 5 adults as one group and 6 children as one group. Now we have 2 groups to arrange.

Step 2: Arrange the two groups

Number of ways to arrange 2 groups: \( 2! \).

Step 3: Arrange within adult group

Number of ways to arrange 5 adults: \( 5! \).

Step 4: Arrange within child group

Number of ways to arrange 6 children: \( 6! \).

Step 5: Total arrangements

Multiply the number of ways for each step: \( 2! \times 5! \times 6! \).
\( 2! = 2 \), \( 5! = 120 \), \( 6! = 720 \), so \( 2\times120\times720 = 172800 \)

Answer:

\( 39916800 \)

Part (b): Alternated (Children and Adults)