Question

The number of ways of arranging the letters AAAAA, BBB, CCC, D, EE & F in a row if the letter C is separated from one another is

$A.{}^{13}{C_3} \times \dfrac{{12!}}{{5!3!2!}} \\\ B.\dfrac{{13!}}{{5!3!2!}} \\\ C.\dfrac{{14!}}{{3!2!3}} \\\ D.{\text{none}} \\\$

Answer 1

Here, we’ll be using the factorial method. In mathematics, factorial is the product of all the positive integers less than or equal to the given positive integer and is denoted by that integer and an exclamation point. Thus, factorial seven is written 7! as 1 × 2 × 3 × 4 × 5 × 6 × 7.
The Zero factorial is defined as equal to the 1. A combination is used if certain objects are to be arranged in such a way that the order of objects is not important.
$^nc{}_r = \dfrac{{n!}}{{r!(n - r)!}}$

Complete step-by-step solution:
Total number of letters of:
A is 5; B is 3; C is 3; D is 1; E is 2 and F is 1​.

According to the question, the arrangement of the letters should be such that no C’s should come together so, $C\\_C\\_C$

In total of 15 letters are there in which no C’s should be together so, there are 12 ways to keep C in between any of the letters as: \\_1\\_2\\_3\\_4\\_5\\_6\\_7\\_8\\_9\\_10\\_11\\_12\\_
The, the probability of arranging C in 13 spaces can be calculated as:
$P = {}^{13}{C_3} \times \dfrac{{12!}}{{5!3!2!}} - - - - (i)$
Here, 5! Is due to 5 A’s, 3! Is due to 3 B’s and 2! is due to 2 E’s. Moreover, ${}^{13}{C_3}$ is due to 3 C’s which can be interchanged within them.
Hence, the number of ways of arranging the letters AAAAA, BBB, CCC, D, EE & F in a row if the letter C is separated from one another is ${}^{13}{C_3} \times \dfrac{{12!}}{{5!3!2!}}$.

Option (A) is the correct answer

Note: Students often got confused with the terms Combination and Permutation. The term Combinations is used if certain objects are to be arranged in such a way that the order of objects is important whereas permutation is the method of selection of the combination, the sequence or the order is utmost important and it is an act of arranging the objects or numbers in the specific order. Mathematically, the formulae for the combination is $^nc{}_r = \dfrac{{n!}}{{r!(n - r)!}}$ and for the permutation is ${}^np{}_r = \dfrac{{n!}}{{(n - r)!}}$. Here, if we want to further evaluate the result then,
${}^{13}{C_3}$ can be evaluated as:
${}^{13}{C_3} = \dfrac{{13!}}{{(13 - 3)!3!}} \\\ = \dfrac{{13!}}{{10!3!}} \\\ = \dfrac{{13 \times 12 \times 11}}{{3 \times 2}} \\\ = 13 \times 11 \times 2 - - - - (ii) \\\$
Now, the value of P is calculated by using equation (i) and (ii) as:

$$P = {}^{13}{C_3} \times \dfrac{{12!}}{{5!3!2!}} \\\ = \left( {13 \times 11 \times 2} \right) \times \dfrac{{12!}}{{5!3!2!}} \\\ = 11 \times \dfrac{{12! \times 13}}{{5!3!}} \\\ = 11 \times \dfrac{{13!}}{{5!3!}} \\\$$