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?

  • A 640
  • B 840
  • C 520
  • D 920

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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