Evaluate \( { }_{8} \mathrm{P}_{5} \) and \( { }_{10} \mathrm{C}_{4} \)

To evaluate \( { }_{8} \mathrm{P}_{5} \), we use the formula for permutations, which is given by: \[ { }_{n} \mathrm{P}_{r} = \frac{n!}{(n-r)!} \] Substituting \( n = 8 \) and \( r = 5 \), we get: \[ { }_{8} \mathrm{P}_{5} = \frac{8!}{(8-5)!} = \frac{8!}{3!} \] Now calculating the factorials: \[ 8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40320 \] \[ 3! = 3 \times 2 \times 1 = 6 \] Thus, \[ { }_{8} \mathrm{P}_{5} = \frac{40320}{6} = 6720 \] Next, to evaluate \( { }_{10} \mathrm{C}_{4} \), we use the combination formula: \[ { }_{n} \mathrm{C}_{r} = \frac{n!}{r!(n-r)!} \] Substituting \( n = 10 \) and \( r = 4 \), we have: \[ { }_{10} \mathrm{C}_{4} = \frac{10!}{4!(10-4)!} = \frac{10!}{4!6!} \] Calculating the necessary factorials: \[ 10! = 10 \times 9 \times 8 \times 7 \times 6! \] This allows us to cancel \( 6! \) in the numerator and denominator: \[ { }_{10} \mathrm{C}_{4} = \frac{10 \times 9 \times 8 \times 7}{4!} \] Calculating \( 4! \): \[ 4! = 4 \times 3 \times 2 \times 1 = 24 \] Then we get: \[ { }_{10} \mathrm{C}_{4} = \frac{10 \times 9 \times 8 \times 7}{24} \] Calculating the numerator: \[ 10 \times 9 = 90 \] \[ 90 \times 8 = 720 \] \[ 720 \times 7 = 5040 \] Finally, performing the division: \[ { }_{10} \mathrm{C}_{4} = \frac{5040}{24} = 210 \] In summary, the values are: \[ { }_{8} \mathrm{P}_{5} = 6720 \] \[ { }_{10} \mathrm{C}_{4} = 210 \]