Math  /  Discrete

QuestionpaCambridge CHAPTER 2. PERMITATIONS AND COMGHINATIONS
Four letters are selected from the 9 letiers in the word CROCODILE. Find the number of selections in which the number of Cs is not the same as the number of O s. [3] \qquad \qquad \qquad

Studdy Solution

STEP 1

1. The word "CROCODILE" consists of the letters C, R, O, C, O, D, I, L, E.
2. The problem requires us to select 4 letters from these 9 letters.
3. We need to count the selections where the number of Cs is not the same as the number of Os.
4. We will use combinatorial methods to determine the number of selections.

STEP 2

1. Count the total possible selections of 4 letters from the word "CROCODILE".
2. Count the selections where the number of Cs is equal to the number of Os.
3. Subtract the number of selections where the number of Cs is equal to the number of Os from the total number of selections.

STEP 3

First, identify the unique letters and their frequencies in the word "CROCODILE".
The word "CROCODILE" consists of: - 2 Cs - 2 Os - 1 R - 1 D - 1 I - 1 L - 1 E

STEP 4

Calculate the total number of ways to choose 4 letters from 9 letters, considering the frequencies (C = 2, O = 2, and each of R, D, I, L, E = 1).
The total number of combinations can be found using the multinomial coefficient.
Total selections=a+b+c+d+e+f+g=40a20b20c,d,e,f,g14!a!b!c!d!e!f!g! \text{Total selections} = \sum_{\substack{a + b + c + d + e + f + g = 4 \\ 0 \leq a \leq 2 \\ 0 \leq b \leq 2 \\ 0 \leq c, d, e, f, g \leq 1}} \frac{4!}{a! b! c! d! e! f! g!}

STEP 5

Let us now count the number of ways to select 4 letters such that the number of Cs is equal to the number of Os.
Possible cases:
1. 1 C and 1 O
2. 2 Cs and 2 Os

We need to calculate the number of selections for each case.

STEP 6

Case 1: Selecting 1 C and 1 O. We need to choose 2 more letters from the remaining 7 letters (R, D, I, L, E, and the remaining C and O).
(21)×(21)×(72)=2×2×21=84 \binom{2}{1} \times \binom{2}{1} \times \binom{7}{2} = 2 \times 2 \times 21 = 84

STEP 7

Case 2: Selecting 2 Cs and 2 Os. There are no additional letters to choose since the selection is already complete.
(22)×(22)=1×1=1 \binom{2}{2} \times \binom{2}{2} = 1 \times 1 = 1

STEP 8

Add the number of selections from each case where the number of Cs is equal to the number of Os.
Equal number of Cs and Os=84+1=85 \text{Equal number of Cs and Os} = 84 + 1 = 85

STEP 9

Calculate the total number of selections of 4 letters from 9 letters by enumerating all possible combinations.
Total selections=a+b+c+d+e+f+g=40a20b20c,d,e,f,g14!a!b!c!d!e!f!g!=126 \text{Total selections} = \sum_{\substack{a + b + c + d + e + f + g = 4 \\ 0 \leq a \leq 2 \\ 0 \leq b \leq 2 \\ 0 \leq c, d, e, f, g \leq 1}} \frac{4!}{a! b! c! d! e! f! g!} = 126

STEP 10

Subtract the number of selections where the number of Cs is equal to the number of Os from the total number of selections.
Selections where number of Cs is not equal to number of Os=12685=41 \text{Selections where number of Cs is not equal to number of Os} = 126 - 85 = 41
Solution: The number of selections in which the number of Cs is not the same as the number of Os is 41 41 .

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