Which of the following is/ are NOT INCORRECT

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There are 12 persons sitting around a circular table. Then the number of ways of selecting 3 out of them such that no two persons are adjacent to each other is 112

If $A = {1, 2, 3}, B = {1, 2, 3, 4, 5, 6, 7, 8}$ then the number of functions ‘f’ from A to B such that $f (i) \leq f (j)$ whenever $i < j$ is 120

A positive integer n is of the form $n = 2^{α} 3^{β},$ where $α \geq 1, β \geq 1$ . If n has 12 positive divisors and 2n has 15 positive divisors, then the number of positive divisors of 3n is 16

Let N be the number of ordered 6-tuples $(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, x_{6})$ of positive integers
satisfying $x_{1} + x_{2} + x_{3} + 3 x_{4} + 3 x_{5} + 5 x_{6} = 21$ , then $\sqrt{N}$ is equal to 9

answer is A, B, C, D.

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Detailed Solution

A) Total number of ways of selecting 3 out of 12 is $12_{c_{3}} = 220$
All these together (123, 234,….,(11, 12, 1)) $\to$ 12 ways
$\begin{matrix} 2 t o g e t h e r a r e 1 a w a y \\ 12 45....11 (8) \\ 23 56....12 (8) \end{matrix}} \to 12 \times 8 = 96$
Required number of arrangements = 220 – 12 – 96 = 112.
B) $n + m - 1_{C_{m}} =^{10} C_{3} = \frac{10 \times 9 \times 8}{6} = 120$

C) Hint: $n = 2^{α} \cdot 3^{β}$
no. of divisors $= (α + 1) (β + 1) = 12$ ... (i)
$n = 2^{α + 1} 3^{β}$ No. of divisors $= (α + 2) (β + 1) = 15$ ... (ii)
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No. of divisors = (3 + 1) (3 + 1) = 16
D) $x_{1} + x_{2} + x_{3} + 3 x_{4} + 3 x_{5} + 5 x_{6} = 21$
Required number of ways = the coefficient of $x^{21}$ in
${(x + x^{2} + x^{3} ...... x^{21})}^{3} {(x^{3} + x^{6} + ..... + x^{21})}^{2} (x^{5} + x^{10} + ..... x^{20}) = 81$