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Question: In how many ways can a 5 letter word be formed from the letters of the word "INDEPENDENCE"...

In how many ways can a 5 letter word be formed from the letters of the word "INDEPENDENCE"

Answer

3345

Explanation

Solution

To form a 5-letter word from the letters of "INDEPENDENCE", we first analyze the available letters and their frequencies:

  • I: 1
  • N: 3
  • D: 2
  • E: 4
  • P: 1
  • C: 1

Total distinct letters: 6 (I, N, D, E, P, C)

We need to consider all possible patterns of repetition for a 5-letter word:

Case 1: All 5 letters are distinct.

  • We need to choose 5 distinct letters from the 6 available distinct letters {I, N, D, E, P, C}.
  • Number of ways to choose 5 distinct letters = 6C5=6^6C_5 = 6.
  • For each set of 5 distinct letters, they can be arranged in 5!5! ways.
  • Number of arrangements = 6×5!=6×120=7206 \times 5! = 6 \times 120 = 720.

Case 2: 4 letters are same, 1 is distinct (e.g., AAAA B).

  • Only 'E' has a frequency of 4. So, the 4 same letters must be E's.
  • The remaining 1 distinct letter (B) must be chosen from {I, N, D, P, C} (5 distinct letters).
  • Number of ways to choose B = 5C1=5^5C_1 = 5.
  • For each such set (EEEE B), the number of arrangements = 5!4!1!=5\frac{5!}{4!1!} = 5.
  • Number of arrangements = 5×5=255 \times 5 = 25.

Case 3: 3 letters are same, 2 are distinct (e.g., AAA BC).

  • The letter appearing 3 times (A) can be 'N' (frequency 3) or 'E' (frequency 4).

    • Subcase 3a: 3 N's.

      • We need to choose 2 distinct letters (B, C) from {I, D, E, P, C} (5 distinct letters, excluding N).
      • Number of ways to choose B, C = 5C2=10^5C_2 = 10.
      • For each set (NNN BC), the number of arrangements = 5!3!1!1!=1206=20\frac{5!}{3!1!1!} = \frac{120}{6} = 20.
      • Number of arrangements = 10×20=20010 \times 20 = 200.

    • Subcase 3b: 3 E's.

      • We need to choose 2 distinct letters (B, C) from {I, N, D, P, C} (5 distinct letters, excluding E).
      • Number of ways to choose B, C = 5C2=10^5C_2 = 10.
      • For each set (EEE BC), the number of arrangements = 5!3!1!1!=1206=20\frac{5!}{3!1!1!} = \frac{120}{6} = 20.
      • Number of arrangements = 10×20=20010 \times 20 = 200.

  • Total for Case 3 = 200+200=400200 + 200 = 400.

Case 4: 3 letters are same, 2 other letters are same (e.g., AAA BB).

  • The letter appearing 3 times (A) can be 'N' or 'E'. The letter appearing 2 times (B) can be 'D', 'N', or 'E'.

    • Subcase 4a: 3 N's and 2 D's.

      • Set of letters: (NNN DD).
      • Number of arrangements = 5!3!2!=1206×2=10\frac{5!}{3!2!} = \frac{120}{6 \times 2} = 10.

    • Subcase 4b: 3 N's and 2 E's.

      • Set of letters: (NNN EE).
      • Number of arrangements = 5!3!2!=1206×2=10\frac{5!}{3!2!} = \frac{120}{6 \times 2} = 10.

    • Subcase 4c: 3 E's and 2 D's.

      • Set of letters: (EEE DD).
      • Number of arrangements = 5!3!2!=1206×2=10\frac{5!}{3!2!} = \frac{120}{6 \times 2} = 10.

    • Subcase 4d: 3 E's and 2 N's.

      • Set of letters: (EEE NN).
      • Number of arrangements = 5!3!2!=1206×2=10\frac{5!}{3!2!} = \frac{120}{6 \times 2} = 10.

  • Total for Case 4 = 10+10+10+10=4010 + 10 + 10 + 10 = 40.

Case 5: 2 letters are same, 2 other letters are same, 1 is distinct (e.g., AA BB C).

  • The letters appearing twice can be 'N', 'D', 'E'. We need to choose two distinct pairs.

    • Subcase 5a: 2 N's and 2 D's.

      • The distinct letter (C) must be chosen from {I, E, P, C} (4 distinct letters, excluding N and D).
      • Number of ways to choose C = 4C1=4^4C_1 = 4.
      • For each set (NN DD C), the number of arrangements = 5!2!2!1!=1202×2=30\frac{5!}{2!2!1!} = \frac{120}{2 \times 2} = 30.
      • Number of arrangements = 4×30=1204 \times 30 = 120.

    • Subcase 5b: 2 N's and 2 E's.

      • The distinct letter (C) must be chosen from {I, D, P, C} (4 distinct letters, excluding N and E).
      • Number of ways to choose C = 4C1=4^4C_1 = 4.
      • For each set (NN EE C), the number of arrangements = 5!2!2!1!=1202×2=30\frac{5!}{2!2!1!} = \frac{120}{2 \times 2} = 30.
      • Number of arrangements = 4×30=1204 \times 30 = 120.

    • Subcase 5c: 2 D's and 2 E's.

      • The distinct letter (C) must be chosen from {I, N, P, C} (4 distinct letters, excluding D and E).
      • Number of ways to choose C = 4C1=4^4C_1 = 4.
      • For each set (DD EE C), the number of arrangements = 5!2!2!1!=1202×2=30\frac{5!}{2!2!1!} = \frac{120}{2 \times 2} = 30.
      • Number of arrangements = 4×30=1204 \times 30 = 120.

  • Total for Case 5 = 120+120+120=360120 + 120 + 120 = 360.

Case 6: 2 letters are same, 3 are distinct (e.g., AA BCD).

  • The letter appearing twice (A) can be 'N', 'D', or 'E'.

    • Subcase 6a: 2 N's.

      • We need to choose 3 distinct letters (B, C, D) from {I, D, E, P, C} (5 distinct letters, excluding N).
      • Number of ways to choose B, C, D = 5C3=10^5C_3 = 10.
      • For each set (NN BCD), the number of arrangements = 5!2!1!1!1!=1202=60\frac{5!}{2!1!1!1!} = \frac{120}{2} = 60.
      • Number of arrangements = 10×60=60010 \times 60 = 600.

    • Subcase 6b: 2 D's.

      • We need to choose 3 distinct letters (B, C, D) from {I, N, E, P, C} (5 distinct letters, excluding D).
      • Number of ways to choose B, C, D = 5C3=10^5C_3 = 10.
      • For each set (DD BCD), the number of arrangements = 5!2!1!1!1!=1202=60\frac{5!}{2!1!1!1!} = \frac{120}{2} = 60.
      • Number of arrangements = 10×60=60010 \times 60 = 600.

    • Subcase 6c: 2 E's.

      • We need to choose 3 distinct letters (B, C, D) from {I, N, D, P, C} (5 distinct letters, excluding E).
      • Number of ways to choose B, C, D = 5C3=10^5C_3 = 10.
      • For each set (EE BCD), the number of arrangements = 5!2!1!1!1!=1202=60\frac{5!}{2!1!1!1!} = \frac{120}{2} = 60.
      • Number of arrangements = 10×60=60010 \times 60 = 600.

  • Total for Case 6 = 600+600+600=1800600 + 600 + 600 = 1800.

Total number of ways:

Sum of arrangements from all cases:

720(Case 1)+25(Case 2)+400(Case 3)+40(Case 4)+360(Case 5)+1800(Case 6)720 (\text{Case 1}) + 25 (\text{Case 2}) + 400 (\text{Case 3}) + 40 (\text{Case 4}) + 360 (\text{Case 5}) + 1800 (\text{Case 6})

=720+25+400+40+360+1800=3345= 720 + 25 + 400 + 40 + 360 + 1800 = 3345.