Question:

<p><strong>Solution:</strong></p><p>Since order of groups is not important, the number of ways to divide 12 different balls into three groups of 5, 4 and 3 balls is:</p><p>$$\frac{12!}{5! \cdot 4! \cdot 3!} = \frac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5!}{5! \cdot (4 \cdot 3 \cdot 2 \cdot 1) \cdot (3 \cdot 2 \cdot 1)} = 27720$$</p><p><strong>Alternate Method:</strong> First group gets 5 balls from 12 in $\binom{12}{5}$ ways. Second group gets 4 balls from remaining 7 in $\binom{7}{4}$ ways. Third group gets 3 balls from remaining 3 in $\binom{3}{3}$ ways.</p><p>Since order of groups is not important:</p><p>$$\binom{12}{5} \times \binom{7}{4} \times \binom{3}{3} = \frac{12!}{5! \cdot 4! \cdot 3!} = 27720$$</p>