Permutations & Combinations
Grade None
Question:
<p>There are 4 girls and 6 boys to be seated such that no two girls sit together. In how many ways can it be done?</p>
<p style="display:inline">302400</p>
<p style="display:inline">302800</p>
<p style="display:inline">604800</p>
<p style="display:inline">60480</p>
Step-by-Step Solution
Key Concept: Use the Gap Method by arranging the unrestricted boys first to create $n+1$ available slots, then place the girls into these gaps to ensure separation.
<p>6 boys can sit in a row in <sup>6</sup>P<sub>6</sub> = 6! ways<br />
— B — B — B — B — B — B —<br />
There are 7 places for 4 girls to sit separately<br />
The number of ways to arrange 4 girls on 7 places = <sup>7</sup>P<sub>4</sub> = 840 ways<br />
The required number of ways = 6! <span class="math-tex">\(\times\)</span> 840 = 720 <span class="math-tex">\(\times\)</span> 840 = 604800</p>
Correct Answer: C