[tex] \large \mathcal{SOLUTION:} [/tex]
[tex] \small \begin{array}{l} \textsf{Recall the formula for combination: } \\ \\ \qquad \large {}^nC_r = \dfrac{n!}{r!\:(n - r)!} \\ \\ \\ a.)\: \bold{Given:}\: {}^nC_5 = 126 \\ \\ \dfrac{n!}{5!\: (n - 5)!} = 126 \\ \\ \textsf{Expand the numerator.} \\ \\ \dfrac{n(n - 1)(n - 2)(n - 3)(n - 4)\cancel{(n - 5)!}}{5!\: \cancel{(n - 5)!}} = 126 \\ \\ \textsf{Multiply both sides by }5! \\ \\ n(n - 1)(n - 2)(n - 3)(n - 4) = 5!(126) \\ \\ n(n - 1)(n - 2)(n - 3)(n - 4) = (5)(4)(3)(2)(1)(3)(6)(7) \\ \\ n(n - 1)(n - 2)(n - 3)(n - 4) = (9)(8)(7)(6)(5) \\ \\ n(n - 1)(n - 2)(n - 3)(n - 4) = (9)(9-1)(9-2)(9-3)(9-4) \\ \\ \therefore \boxed{n = 9} \\ \\ \: \end{array} [/tex]
[tex] \small \begin{array}{l} b.)\: \bold{Given:}\: {}^{10}C_r = 252 \\ \\ \begin{aligned} \dfrac{10!}{r! \: (10 - r)!} = 252 \implies \dfrac{10!}{252} &= r! (10 - r)! \\ \\ \dfrac{3628800}{252} &= r! (10 - r)! \\ \\ 1440 &= r!(10 - r)! \\ \\ (120)^2 &= r!(10 - r)! \\ \\ (10 - r)! &= r! = 120 \\ \\ \therefore &\:\:\boxed{r = 5} \end{aligned} \end{array} [/tex]
