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What is the total number of different 11-letter arrangements that can be formed using the letters in the word GALVANIZING?
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What is the total number of different 1111-letter arrangements that can be formed using the letters in the word GALVANIZING?\newlineAnswer:

Full solution

Q. What is the total number of different 1111-letter arrangements that can be formed using the letters in the word GALVANIZING?\newlineAnswer:
  1. Word Analysis: The word GALVANIZING has 1111 letters with the following counts of identical letters: G2G-2, A2A-2, N2N-2, L1L-1, V1V-1, I1I-1, Z1Z-1.
  2. Permutations Formula: To find the total arrangements, we use the formula for permutations of a multiset: n!n1!×n2!××nk!\frac{n!}{n_1! \times n_2! \times \ldots \times n_k!}, where nn is the total number of items to arrange, and n1,n2,,nkn_1, n_2, \ldots, n_k are the counts of identical items.
  3. Calculate 11!11!: So, we calculate 11!(2!2!2!)\frac{11!}{(2! \cdot 2! \cdot 2!)} to account for the repeated G's, A's, and N's.
  4. Calculate 2!2!: Calculating 11!11! gives us 3991680039916800.
  5. Calculate Total Sets: Calculating 2!2! for one set of identical letters gives us 22.
  6. Divide to Get Total: Since we have three sets of identical letters, we calculate 2!×2!×2!2! \times 2! \times 2! which equals 88.
  7. Divide to Get Total: Since we have three sets of identical letters, we calculate 2!×2!×2!2! \times 2! \times 2! which equals 88.Now, we divide 3991680039916800 by 88 to get the total number of different arrangements.
  8. Divide to Get Total: Since we have three sets of identical letters, we calculate 2!×2!×2!2! \times 2! \times 2! which equals 88.Now, we divide 3991680039916800 by 88 to get the total number of different arrangements.39916800÷839916800 \div 8 equals 49896004989600.

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