251.

The set S: {1, 2, 3, …, 12} is to be partitioned into three sets A, B, C of equal size. Thus, A ∪ B ∪ C = S, A ∩ B = B ∩ C = A ∩ C = φ. The number of ways to partition S is-

• 12!/3!(4!)³

• 12!/3!(3!)⁴

• 12!/(4!)³

• 12!/(4!)³

The value  of 

How many ways are there to arrange the letters in the word GARDEN with the vowels in alphabetical order?

• 120

• 480

• 360

• 360

The number of ways of distributing 8 identical balls in 3 distinct boxes so that none of the boxes is empty is

• 5

• ⁸C₃

• 3⁸

• 3⁸

The number of all numbers having 5 digits, with distinct digits is

• 99999

• $9 \times {}^9\mathrm{P}_4$

• ${}^{10}\mathrm{P}_5$

• ${}^9\mathrm{P}_4$

The greatest integer which divides (p + 1) (p + 2) (p + 3) .... (p + q) for all p E N and fixed q $\in$ N is

• p!

• q!

• p

• q

Let ${\left(1 +\hspace{0.17em}\mathrm{x} + {\mathrm{x}}^2\right)}^9 = {\mathrm{a}}_0 +\hspace{0.17em}{\mathrm{a}}_1\mathrm{x} +\hspace{0.17em}{\mathrm{a}}_2{\mathrm{x}}^2 +\hspace{0.17em}... + {\mathrm{a}}_{18}{\mathrm{x}}^{18}$. Then,

• ${\mathrm{a}}_0 +\hspace{0.17em}{\mathrm{a}}_2 +\hspace{0.17em}... +\hspace{0.17em}{\mathrm{a}}_{18} = {\mathrm{a}}_0 +\hspace{0.17em}{\mathrm{a}}_3 + ... + {\mathrm{a}}_{17}$

• ${\mathrm{a}}_0 + {\mathrm{a}}_2 +\hspace{0.17em}... + {\mathrm{a}}_{18}$ is even

• ${\mathrm{a}}_0 + {\mathrm{a}}_2 +\hspace{0.17em}... + {\mathrm{a}}_{18}$ is divisible by 9

• ${\mathrm{a}}_0 + {\mathrm{a}}_2 +\hspace{0.17em}... + {\mathrm{a}}_{18}$ is divisible by 3 but not by 9

The number of ways in which the letters of the word ARRANGE can be permuted such that the R's occur together, is

• $\frac{7!}{2! 2!}$

• $\frac{7!}{2!}$

• $\frac{6!}{2!}$

• $5! \times 2!$

If $\frac{1}{{}^5\mathrm{C}_{\mathrm{r}}} + \frac{{\displaystyle 1}}{{\displaystyle {}^6\mathrm{C}_{\mathrm{r}}}} = \frac{{\displaystyle 1}}{{\displaystyle {}^4\mathrm{C}_{\mathrm{r}}}}$, then the value of r is

• 4

• 2

• 5

• 3

If A = $\left\{5^{\mathrm{n}} - 4\mathrm{n} - 1: \mathrm{n} \in \mathrm{N}\right\}$ and $\mathrm{B} = \left\{16\left(\mathrm{n} - 1: \mathrm{n} \in \mathrm{N}\right)\right\}$, then

• A = B

• $\mathrm{A} \cap \mathrm{B} = \mathrm{\phi }$

• $\mathrm{A} \subseteq \mathrm{B}$

• $\mathrm{B} \subseteq \hspace{0.17em}\mathrm{A}$