Evaluate the expression. a) \( { }_{7} \mathrm{C}_{5} \) b) \( { }_{7} \mathrm{P}_{5} \)

**Step-by-step solution:** 1. **Evaluate \( { }_{7}\mathrm{C}_{5} \):** The formula for combinations is: \[ { }_{n}\mathrm{C}_{r} = \frac{n!}{r!(n-r)!} \] For \( n = 7 \) and \( r = 5 \): \[ { }_{7}\mathrm{C}_{5} = \frac{7!}{5!(7-5)!} = \frac{7!}{5! \cdot 2!} \] Calculate \( 7! \), \( 5! \), and \( 2! \): \[ 7! = 5040,\quad 5! = 120,\quad 2! = 2 \] Substitute these values: \[ { }_{7}\mathrm{C}_{5} = \frac{5040}{120 \cdot 2} = \frac{5040}{240} = 21 \] 2. **Evaluate \( { }_{7}\mathrm{P}_{5} \):** The formula for permutations is: \[ { }_{n}\mathrm{P}_{r} = \frac{n!}{(n-r)!} \] For \( n = 7 \) and \( r = 5 \): \[ { }_{7}\mathrm{P}_{5} = \frac{7!}{(7-5)!} = \frac{7!}{2!} \] Using the previously computed values: \[ { }_{7}\mathrm{P}_{5} = \frac{5040}{2} = 2520 \] **Final Answers:** a) \( { }_{7}\mathrm{C}_{5} = 21 \) b) \( { }_{7}\mathrm{P}_{5} = 2520 \)