Permutation of Alike Items Explained with Examples

Understanding Permutations of Alike Items

Permutations refer to the arrangement of objects in a specific order. When all objects are distinct, the number of linear arrangements of n objects is n! (n factorial). However, when some objects are identical (alike), we divide by the factorials of the frequencies of the repeated items to avoid overcounting.

The formula for the number of distinct linear arrangements of n items, where p items are alike of one kind, q items are alike of another kind, r items are alike of a third kind, and the rest are distinct, is:

\[ \frac{n!}{p! \cdot q! \cdot r!} \]

Here, \( p + q + r + \text{(rest)} = n \).

Example Problems and Solutions

Question 1: How many 12-letter words can be formed using the letters in the word “MISSISSIPPII”?

Solution:

Step 1: Identify total letters and frequencies:

  • Total letters (n) = 12
  • I appears 5 times
  • S appears 4 times
  • P appears 2 times
  • M appears 1 time

Step 2: Apply the formula:

\[ \frac{12!}{5! \cdot 4! \cdot 2! \cdot 1!} \]

Calculation:

  • 12! = 479,001,600
  • 5! = 120, 4! = 24, 2! = 2, 1! = 1
  • Denominator = 120 × 24 × 2 × 1 = 5,760

Number of arrangements = 479,001,600 / 5,760 = 83,160

Answer: 83,160 words

Question 2: How many 6-digit numbers can be formed using the digits 1, 3, 5, 7 if one digit is repeated exactly three times?

Solution:

Method 1 (Case-by-case):

\[ \frac{6!}{3!} = 120 \]

There are 4 possible digits to repeat → Total = 4 × 120 = 480

Method 2 (Combinatorial):

\[ {}^4C_1 \times \frac{6!}{3!} = 4 \times 120 = 480 \]

Answer: 480 numbers

Question 3: How many 6-digit numbers can be formed using the digits 1, 3, 5, 7 if there are two pairs of digits?

Solution:

Method 1 (Case-by-case):

\[ \frac{6!}{2! \cdot 2!} = 180 \]

There are 6 possible pairs → Total = 6 × 180 = 1,080

Method 2 (Combinatorial):

\[ {}^4C_2 \times \frac{6!}{2! \cdot 2!} = 6 \times 180 = 1,080 \]

Answer: 1,080 numbers

Question 4: How many 7-letter words can be formed using the letters in “BANANA”?

Solution:

  • Total letters (n) = 7
  • A appears 3 times
  • N appears 2 times
  • B appears 1 time
\[ \frac{7!}{3! \cdot 2! \cdot 1!} = \frac{5040}{6 \cdot 2 \cdot 1} = 420 \]

Answer: 420 words

Question 5: How many 5-digit numbers can be formed using the digits 2, 4, 6 if one digit is repeated exactly four times?

Solution:

\[ {}^3C_1 \times {}^2C_1 \times \frac{5!}{4!} = 3 \times 2 \times 5 = 30 \]

Answer: 30 numbers

Conclusion

Permutations of alike items are an important concept in combinatorics. The key is to divide n! by the factorials of the counts of repeated items. Regular practice with such problems helps build strong problem-solving skills in Engineering Mathematics.

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