Question

If $^{n}P_{5}=6720$ and $^{(n-1)}P_{4}=840$, then $^{n}P_{3}$ is equal to

• A. 346
• B. 348
• C. 396
• D. 376
• E. 336

Hint:

Math Tip: Expanding factorials completely is slow. Remembering the recursive relation $^nP_r = n \cdot ^{(n-1)}P_{(r-1)}$ turns a complex factorial division problem into a simple single-step linear equation.

Answer 1

Answer: (E)

Concept:
Combinatorics - Permutations.
A key property of permutations linking consecutive values is: $^nP_r = n \times ^{(n-1)}P_{(r-1)}$.
Step 1: Apply the permutation property to the given equations.
Using the property $^nP_r = n \times ^{(n-1)}P_{(r-1)}$, substitute $r = 5$: $$ ^nP_5 = n \times ^{(n-1)}P_4 $$
Step 2: Substitute the known values to find n.
We are given $^nP_5 = 6720$ and $^{(n-1)}P_4 = 840$. $$ 6720 = n \times 840 $$
Step 3: Solve for n.
Divide both sides by 840: $$ n = \frac{6720}{840} $$ $$ n = 8 $$
Step 4: Set up the equation for the target expression.
Now that we know $n = 8$, substitute it into the expression we need to find, which is $^nP_3$. $$ ^nP_3 = ^8P_3 $$
Step 5: Calculate the final value.
Evaluate $^8P_3$ by taking the first 3 terms of the descending factorial of 8: $$ ^8P_3 = 8 \times 7 \times 6 $$ $$ = 56 \times 6 $$ $$ = 336 $$