Five letters are selected at random from the 9 letter of the word INFLUENCE. find the number of different selections if the 5 letter include at least one N and at most one E.

To solve this problem, we need to find the number of different ways to select 5 letters from the 9-letter word "INFLUENCE" such that the selection includes at least one "N" and at most one "E".

### Step-by-Step Solution

1. **Count the total number of each letter in the word "INFLUENCE":**

   - **I**: 1
   - **N**: 2
   - **F**: 1
   - **L**: 1
   - **U**: 1
   - **E**: 2
   - **C**: 1

   We have 9 letters in total: I, N, N, F, L, U, E, E, C.

2. **Conditions for selection:**
   - At least one "N"
   - At most one "E"

3. **Break down the problem based on the number of "E" letters selected:**
   We can select either 0 or 1 "E" (since at most one "E" is allowed).

   **Case 1: Select 0 E's**

   - We are left with the letters: I, N, N, F, L, U, C (7 letters).
   - We need to select 5 letters from these 7 letters, including at least one "N".
   
   Let’s first count the total number of ways to select 5 letters from 7:
   \[
   \binom{7}{5} = 21.
   \]

   Now, subtract the cases where no "N" is selected. If no "N" is selected, we choose 5 letters from the remaining 5 letters (I, F, L, U, C):
   \[
   \binom{5}{5} = 1.
   \]

   Therefore, the number of ways to select 5 letters from these 7 with at least one "N" and no "E" is:
   \[
   21 - 1 = 20.
   \]

   **Case 2: Select 1 E**

   - We select 1 "E", so we are left with the letters I, N, N, F, L, U, C (7 letters).
   - Now, we need to select 4 more letters from these 7 letters, including at least one "N".
   
   The total number of ways to select 4 letters from 7:
   \[
   \binom{7}{4} = 35.
   \]

   Again, subtract the cases where no "N" is selected. If no "N" is selected, we choose 4 letters from the remaining 5 letters (I, F, L, U, C):
   \[
   \binom{5}{4} = 5.
   \]

   Therefore, the number of ways to select 4 letters from these 7 with at least one "N" and exactly one "E" is:
   \[
   35 - 5 = 30.
   \]

4. **Total number of selections:**

   Combine the results from both cases:
   \[
   20 + 30 = 50.
   \]

### Conclusion

The number of different selections of 5 letters that include at least one "N" and at most one "E" is **50**.

---

Would you like a deeper explanation of any part, or do you have other questions?

Here are 5 related questions for further exploration:

1. How would the solution change if the condition was "at least two N's" instead?
2. How many ways can we select 5 letters if there were no restrictions?
3. What if the condition was "at least two E's" instead of "at most one E"?
4. How would you calculate the probability that a random selection of 5 letters meets the given criteria?
5. Can you generalize the solution to any word with a similar structure?

**Tip:** When dealing with problems involving restrictions, it's often helpful to calculate the total number of possibilities and then subtract the cases that do not meet the criteria.

Five letters are selected at random from the 9-letter word INFLUENCE. Find the number of different selections if the 5 letters include at least one N and at most one E.

Discover how to count the number of ways to select five letters from the word INFLUENCE with specific restrictions: at least one N and at most one E. This problem involves combinatorics and uses the combination formula, suitable for students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation