Math Problem Statement

4 boys and 3 girls can sit in alternate in such ways answer is

Solution

To find the number of ways in which 4 boys and 3 girls can sit alternately, we will proceed step by step.

Step 1: Total seats arrangement

Since there are 4 boys and 3 girls, alternate seating means the arrangement would either start with a boy or start with a girl. In this case, since there are more boys, the arrangement must start with a boy.

So, the seating pattern will be: Boy, Girl, Boy, Girl, Boy, Girl, Boy\text{Boy, Girl, Boy, Girl, Boy, Girl, Boy}

Step 2: Arranging the boys

There are 4 boys, and they can be arranged in the boy positions (1st, 3rd, 5th, and 7th seats). The number of ways to arrange 4 boys in these positions is: 4!=4×3×2×1=244! = 4 \times 3 \times 2 \times 1 = 24

Step 3: Arranging the girls

There are 3 girls, and they can be arranged in the girl positions (2nd, 4th, and 6th seats). The number of ways to arrange 3 girls in these positions is: 3!=3×2×1=63! = 3 \times 2 \times 1 = 6

Step 4: Total number of arrangements

The total number of ways to arrange the 4 boys and 3 girls alternately is: 4!×3!=24×6=1444! \times 3! = 24 \times 6 = 144

Thus, the total number of ways they can sit alternately is 144.

Would you like further details or explanations on this? Here are some related questions to think about:

  1. What if there were 4 boys and 4 girls? How would the number of arrangements change?
  2. How does the formula for permutations (n!) work, and why do we multiply them for the two groups?
  3. What happens if the number of boys and girls is equal? Can they still sit alternately?
  4. What if we allowed boys or girls to sit together, how many seating arrangements would there be then?
  5. If a specific boy and girl must sit together, how does that affect the number of ways?

Tip:

When solving permutation problems, always check for constraints like "alternating" or "together," as they significantly affect the final answer.

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Math Problem Analysis

Mathematical Concepts

Permutations
Combinatorics

Formulas

n! = n × (n - 1) × ... × 1

Theorems

Factorial Theorem for Arrangements

Suitable Grade Level

Grades 9-11