Question 1 was asked before, but not answered.


1. Find the number of ways to arrange the letters A, B, B, C, C, C. (Both B's are identical, and all three C's are identical.)


2. Find the value of the sum \(\dbinom{13}{0}+\dbinom{13}{1}+\dbinom{13}{2}+\dbinom{13}{3}+\dbinom{13}{4}+\dbinom{13}{5}+\dbinom{13}{6}\)


3. How many 4-digit numbers are there where any two consecutive digits are different?


Thank you for your help!

 Mar 1, 2020

1)  If you had 6 different letters, the answer would be 6!.

     Since you have multiples, you need to divide by the factorials of the multiples:  Answer = 6! / (2! x 3!)

     -- because you have 2 Bs and 3 Cs.


2)  You can find the value of each one individually, and sum them together.

      Or, since this sum gives you half of the entries of the 13th row of Pascal's triangle, you can find the sum of that row

      (which is 213) and divide that  answer by 2.


3)  Since the first digit can be any digit but zero, you have 9 choices for the first digit.

     Since the next digit can be any digit that doesn't match the first digit, you again have 9 choices for the second digit.

     Similar, for digits 3 and 4.

     Total:  9 x 9 x 9 x 9

 Mar 1, 2020

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