UGC NET CSE | December 2012 | Part 3 | Question: 20

Question

The number of ways to distribute n distinguishable objects into k distinguishable boxes, so that $n_i$ objects are placed into box $i$, $i=1, 2, \dots k$ equals which of the following?

• A. $\frac{n!}{n_1!+n_2!+ \dots + n_k!}$
• B. $\frac{ n_1!+n_2!+ \dots + n_k!}{n_1! n_2! \dots  n_k!}$
• C. $\frac{ n_1!}{n_1! n_2! \dots  n_k!}$
• D. $\frac{ n_1! n_2!  \dots  n_k!}{n_1! - n_2! – n_3! \dots - n_k!}$

Answer 1

C is answer

The number of ways to distribute n distinguishable objects into k distinct boxes so that ni objects are placed in box i, i=1, ..., k, and n1+...+nk = n, is 

n₁ !/ n₁ .n₂. n₃.....n_k

Comments

Is it n or n₁ in numerator