Question

All letters of the word BREAKAGE are to be jumbled. The number of ways of arranging them so that no two A's and no two E's are together

• A. 5760
• B. 10080
• C. 4320
• D. 720

Answer 1

Answer: (A)

5760

Answer 2

The problem requires finding arrangements of "BREAKAGE" where "AA" and "EE" do not appear. This is solved using the Principle of Inclusion-Exclusion.

1. Calculate the total number of arrangements of "BREAKAGE": $\frac{8!}{2!2!} = 10080$.
2. Calculate arrangements with "AA" together: Treat "AA" as one unit. Arrange (AA), B, R, E, E, K, G. This gives $\frac{7!}{2!} = 2520$.
3. Calculate arrangements with "EE" together: Treat "EE" as one unit. Arrange B, R, (EE), A, A, K, G. This gives $\frac{7!}{2!} = 2520$.
4. Calculate arrangements with "AA" together AND "EE" together: Treat "AA" and "EE" as units. Arrange (AA), (EE), B, R, K, G. This gives $6! = 720$.
5. Using Inclusion-Exclusion, the number of arrangements with at least one pair together is $2520 + 2520 - 720 = 4320$.
6. Subtract this from the total arrangements to find arrangements with no pairs together: $10080 - 4320 = 5760$.