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Question: Find the number of ways in which the 9 letters in the word CROCODILE can be divided into three group...

Find the number of ways in which the 9 letters in the word CROCODILE can be divided into three groups, each containing three letters, if the two Cs must be in different groups.

A

(A) 15

B

(B) 10

C

(C) 30

D

(D) 55

Answer

55

Explanation

Solution

Here's how to solve this problem:

The word CROCODILE has 9 letters: C (2 times), R (1 time), O (2 times), D (1 time), I (1 time), L (1 time), E (1 time). We want to divide these into three groups of three letters each, with the two C's in different groups.

1. Assign the C's

Since the two C's must be in different groups, we place one C in group 1 and the other C in group 2. Group 3 will not contain a C.

  • Group 1: {C, _, _}
  • Group 2: {C, _, _}
  • Group 3: {_, _, _}

The remaining 7 letters are {R, O, O, D, I, L, E}. We need to distribute these to complete the three groups: 2 letters for Group 1, 2 letters for Group 2, and 3 letters for Group 3.

2. Consider the distribution of O's

We have three cases to consider based on where the two O's are placed:

  • Case 1: Both O's are in Group 3

    • Group 1: {C, a, b}
    • Group 2: {C, c, d}
    • Group 3: {O, O, e}

    The letters {a, b, c, d, e} must be chosen from the remaining 5 distinct letters {R, D, I, L, E}. The number of ways to choose {a, b} for Group 1 is (52)=10\binom{5}{2} = 10. The number of ways to choose {c, d} for Group 2 from the remaining 3 letters is (32)=3\binom{3}{2} = 3. The remaining 1 letter is {e} for Group 3: (11)=1\binom{1}{1} = 1.

    The number of ways to choose the letters in this case is 10×3×1/2!=1510 \times 3 \times 1 / 2! = 15. We divide by 2! because the groups containing C and two other letters are indistinguishable.

  • Case 2: One O is in Group 1 and one O is in Group 2

    • Group 1: {C, O, a}
    • Group 2: {C, O, b}
    • Group 3: {c, d, e}

    The letters {a, b, c, d, e} must be chosen from the remaining 5 distinct letters {R, D, I, L, E}.

    The number of ways to choose {c, d, e} for Group 3 is (53)=10\binom{5}{3} = 10.

  • Case 3: One O is in Group 1 and one O is in Group 3

    • Group 1: {C, O, a}
    • Group 2: {C, b, c}
    • Group 3: {O, d, e}

    The letters {a, b, c, d, e} must be chosen from the remaining 5 distinct letters {R, D, I, L, E}.

    The number of ways to choose {a} for Group 1 is (51)=5\binom{5}{1} = 5. The number of ways to choose {b, c} for Group 2 from the remaining 4 letters is (42)=6\binom{4}{2} = 6.

    The number of ways to choose {d, e} for Group 3 is (22)=1\binom{2}{2} = 1. The number of ways to choose the letters in this case is 5×6×1=305 \times 6 \times 1 = 30.

3. Total Number of Ways

The total number of ways to form the three groups with the C's separated is the sum of the ways in the three cases:

15+10+30=5515 + 10 + 30 = 55

Therefore, the answer is 55.