Evaluate \( { }_{14} \mathrm{C}_{4} \) and \( { }_{14} \mathrm{P}_{4} \). \( 14 \mathrm{C}_{4}=\square \)

We start with the definition of combinations: \[ { }_{n}\mathrm{C}_{r}=\frac{n!}{r!(n-r)!} \] For \( n=14 \) and \( r=4 \): \[ { }_{14}\mathrm{C}_{4}=\frac{14!}{4!(14-4)!}=\frac{14!}{4!10!} \] Notice that the factorial terms from 10! will cancel out with part of 14!: \[ { }_{14}\mathrm{C}_{4}=\frac{14 \times 13 \times 12 \times 11 \times 10!}{4 \times 3 \times 2 \times 1 \times 10!}=\frac{14 \times 13 \times 12 \times 11}{4 \times 3 \times 2 \times 1} \] Calculating the numerator: \[ 14 \times 13=182,\quad 182 \times 12=2184,\quad 2184 \times 11=24024 \] And the denominator: \[ 4 \times 3 \times 2 \times 1=24 \] Thus, \[ { }_{14}\mathrm{C}_{4}=\frac{24024}{24}=1001 \] Next, we use the definition of permutations: \[ { }_{n}\mathrm{P}_{r}=\frac{n!}{(n-r)!} \] For \( n=14 \) and \( r=4 \): \[ { }_{14}\mathrm{P}_{4}=\frac{14!}{(14-4)!}=\frac{14!}{10!} \] Writing the numerator out to the cancellation point: \[ { }_{14}\mathrm{P}_{4}=14 \times 13 \times 12 \times 11 \] We already computed this product: \[ 14 \times 13 \times 12 \times 11 = 24024 \] Therefore, the evaluations are: \[ { }_{14}\mathrm{C}_{4}=1001 \quad \text{and} \quad { }_{14}\mathrm{P}_{4}=24024 \] Finally, the answer for \( { }_{14}\mathrm{C}_{4} \) is: \[ 1001 \]