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The no. of ways to distribute n distinguishable objects into k distinguishable boxes, so that \(n_i\) objects are placed into box i, i = 1, 2, …. k equals which of the following ?

  • A.\(\frac{n!}{n_1! + n_2! + \cdots + n_k!}\)
  • B.\(\frac{n_1! + n_2! + \cdots + n_k!}{n_1!\, n_2!\, n_3! \cdots n_k!}\)
  • C.\(\frac{n!}{n_1!\, n_2!\, n_3! \cdots n_k!}\)
  • D.\(\frac{n_1!\, n_2! \cdots n_k!}{n_1! - n_2! - n_3! \cdots - n_k!}\)

Answer: C

The number of ways to distribute n distinguishable objects into k distinguishable boxes so that box i gets \(n_i\) objects is the multinomial coefficient

\[ \frac{n!}{n_1!\, n_2! \cdots n_k!} \]

The reasoning is that the n objects can be arranged in n! orders, and the order inside each box does not matter, so the count is divided by \(n_i!\) for each box. The options with sums of factorials in the numerator or denominator have no such combinatorial meaning. The keyed answer is the third option.

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