How many distinct permutations can be made from the letters of the word PHILIPPINES?
The correct answer and explanation is :
Let’s analyze the problem step-by-step.
Word: PHILIPPINES
Step 1: Count the total number of letters
The word PHILIPPINES has 10 letters in total.
Step 2: Identify repeated letters and their frequencies
- P appears 3 times
- I appears 3 times
- H appears 1 time
- L appears 1 time
- N appears 1 time
- E appears 1 time
- S appears 1 time
Step 3: Use the formula for permutations of multiset
The total number of distinct permutations of letters in a word with repeated letters is: n!n1!×n2!×⋯\frac{n!}{n_1! \times n_2! \times \cdots}
Where:
- nn = total letters
- n1,n2,…n_1, n_2, \ldots = frequencies of each repeated letter
For PHILIPPINES: n=10n = 10 nP=3,nI=3n_P = 3, \quad n_I = 3
All other letters appear once, so their factorials are 1 and don’t affect the denominator.
Step 4: Calculate factorials and apply the formula
Number of distinct permutations=10!3!×3!\text{Number of distinct permutations} = \frac{10!}{3! \times 3!}
Calculate:
- 10!=3,628,80010! = 3,628,800
- 3!=63! = 6
So, 3,628,8006×6=3,628,80036=100,800\frac{3,628,800}{6 \times 6} = \frac{3,628,800}{36} = 100,800
Final answer:
100,800 distinct permutations can be made from the letters of the word PHILIPPINES.
Explanation (300 words):
When counting permutations of a word with repeated letters, it’s essential to account for indistinguishable arrangements caused by identical letters. The word “PHILIPPINES” contains 10 letters with two sets of repeated letters: three P’s and three I’s.
If all letters were distinct, the total number of permutations would be 10!10!, which equals 3,628,800. This number represents every possible ordering of 10 unique letters.
However, because three P’s are indistinguishable among themselves, swapping these P’s does not create new unique permutations. Similarly, the three I’s are identical and interchanging them also does not produce new distinct permutations. To correct for this overcounting, we divide by the factorial of the counts of each repeated letter.
The number of ways to rearrange the three identical P’s among themselves is 3!=63! = 6, and the same applies for the three I’s. Dividing the total permutations by 3!×3!3! \times 3! removes duplicates that are identical due to these repetitions.
Hence, the formula used is: 10!3!×3!\frac{10!}{3! \times 3!}
This yields: 3,628,8006×6=100,800\frac{3,628,800}{6 \times 6} = 100,800
Therefore, there are 100,800 unique ways to arrange the letters of “PHILIPPINES,” accounting properly for repeated letters. This approach is general and applies to any problem involving permutations of words with repeated letters, ensuring an accurate count of distinct permutations.