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Worked Examples · Example 29

Q.The MASSASAUGA is a brown and white venomous snake found in North America. How many arrangements can be made from this word. Find the number of arrangements if all vowels are together.

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MASSASAUGA has 10 letters (M×1, A×4, S×3, U×1, G×1); total arrangements =10!/(4! 3!)=25200=10!/(4!\,3!)=25200, and with all vowels together =100=100.

Number of distinct arrangements of nn objects with repeated items of counts p1,p2,…p_1,p_2,\dots is

n!p1! p2! ⋯\dfrac{n!}{p_1!\,p_2!\,\cdots}

where nn = total letters and pip_i = frequency of each repeated letter.

  1. List and count letters of MASSASAUGA: M, A, S, S, A, S, A, U, G, A → n=10n=10 letters with M=1=1, A=4=4, S=3=3, U=1=1, G=1=1.
  2. Total arrangements (no restriction):

10!4! 3!=362880024×6=3628800144=25200\dfrac{10!}{4!\,3!}=\dfrac{3628800}{24\times6}=\dfrac{3628800}{144}=25200

  1. Vowels together: vowels are A, A, A, A, U (5 vowels: A×4, U×1); consonants are M, S, S, S (4 letters: M×1, S×3).
  2. Treat the block of 5 vowels as one unit. Together with the 4 consonants, we arrange 4+1=54+1=5 units where S repeats 3 times: 5!3!=1206=20\dfrac{5!}{3!}=\dfrac{120}{6}=20 …

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