A senatorial candidate had planned to visit seven cities prior to a primary election. However, he could only visit four of the cities. How many different itineraries could be considered?
The correct answer is B. 840
Sure, I can help you solve this problem. To find the number of different itineraries, we can use the concept of permutations. The formula for permutations is:
\((P(n, r) = \frac{n!}{(n - r)!} \)
Where:
- \( n \) is the total number of items (cities in this case),
- \( r \) is the number of items to be chosen (cities to be visited), and
- \( ! \) represents the factorial of a number.
In this case, the senatorial candidate had planned to visit seven cities (\( n = 7 \)), but could only visit four (\( r = 4 \)).
Let's plug in the values and calculate:
\( P(7, 4) = \frac{7!}{(7 - 4)!} = \frac{7!}{3!} = \frac{5040}{6} = 840 \)
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