Question

Ques no. 23 & 24 on
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Answer 1

Answer on Question #56700 – Math– Combinatorics | Number Theory

23. Consider $xyz = 24$, where $x, y, z \in I$, then

(a) Total number of positive integral solutions for $x, y, z$ are 81

(b) Total number of positive integral solutions for $x, y, z$ are 90

(c) Total number of positive integral solutions for $x, y, z$ are 30

(d) Total number of positive integral solutions for $x, y, z$ are 120

Solution

Because $24 = 2^3 \cdot 3^1$, then there are 4 prime positive divisors of 24. It follows, that total number of positive integer solutions for $x, y, z$ is $\overline{A_3^4} = 3^4 = 81$. The total number of all integer solutions for $x, y, z$ is $81 \cdot 2^3 = 81 \cdot 8 = 648$.

Answer: (a).

24: If ${}^n C_{r+1} = (m^2 - 8) \cdot {}^{n-1} C_r$; then possible value of ${}'m'$ can be

(a) 4 (b) 2 (c) 3 (d) -5

Solution

We have:

$${}^n C_{r+1} = \frac{n!}{(n-r-1)!(r+1)!} = \frac{n \cdot (n-1)!}{(r+1) \cdot (n-r-1)!r!} = (m^2 - 8) \cdot {}^{n-1} C_r = (m^2 - 8) \cdot \frac{(n-1)!}{(n-r-1)!r!},$$

hence

$$n = (m^2 - 8)(r + 1)$$

and $\frac{n}{r+1}$ must be a positive integer, $1 < \frac{n}{r+1} \leq n$, hence $m^2 - 8 > 1$, that is, $m^2 > 9$.

Thus, $m \neq 2$, $m \neq 3$, and possible value of ${}'m'$ can be 4 (for $n = 16$, $r = 1$, for example) or (-5) (for $n = 17$, $r = 0$, for example).

Answer: (a) or (d).

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