Question

The number of permutations of letters $a,b,c,d,e,f,g$ so that neither the pattern $beg$ nor $cad$ appears is
A.$\dfrac{{7!}}{{3!3!}}$
B.$\dfrac{{7!}}{{2!3!3!}}$
C.$4806$
D.None of these

Answer 1

Here, we have to use the concept of permutation to find the possible number of permutations of letters. So, firstly we will find the total number of possible permutations of letters from all the given letters. Then we will find the total possible number of permutations of letters with the pattern $beg$ and $cad$ then we will subtract these to get our required number of permutations of letters.

Complete step-by-step answer:
There are 7 letters given i.e. $a,b,c,d,e,f,g$
Firstly we will find the total number of possible permutations of letters from all the given letters. Therefore, we get
Total number of possible permutations of letters from all the given letters $= 7! = 5040$… (1)
Now we will find the total possible number of permutations of letters with the pattern $beg$ and $cad$ i.e. $n\left( {A \cup B} \right)$.
Firstly we will consider A to be the function of permutation in which $beg$ occurs and B be the function of permutation in which $cad$ occurs.
So we have to find the $n\left( {A \cup B} \right)$ and we know the formula of $n\left( {A \cup B} \right)$ which is
$n\left( {A \cup B} \right) = n\left( A \right) + n\left( B \right) - n\left( {A \cap B} \right)$
So for the function A we consider $beg$ as a single term then the number of possible permutations will be$5!$ because $beg$ will be treated as the one letter in whole and there are four remaining letters. Therefore
$n\left( A \right) = 5!$
Similarly for the function B, If we consider $cad$ as a single term then the number of possible permutations will be $5!$ because $cad$ will be treated as the one letter in whole and there are four remaining letters. Therefore
$n\left( B \right) = 5!$
Now we have to find$n\left( {A \cap B} \right)$. So for this condition it means that in this permutation both the words $beg$ and $cad$ is present. So if these two words are present in it then there is only one letter remaining. So the number of possible permutations will be $3!$.
Therefore, value of $n\left( {A \cup B} \right)$ is
$n\left( {A \cup B} \right) = 5! + 5! - 3! = 120 + 120 - 6 = 234$
Therefore, the required number of permutations of letters will be the difference between the total number of possible number of permutations of letters from all the given letters and the total possible number of permutations of letters with the pattern $beg$ and $cad$.
Required number of permutations of letters $= 5040 - 234 = 4806$
Hence, option C is the correct option.

Note: Here we have to note that we need to apply the concept of permutation not the combination for the arrangement of the letters. As a permutation is defined as the different ways in which a collection of items can be arranged. For example: The different ways in which the numbers 1, 2 and 3 can be grouped together, taken all at a time, are$123,{\rm{ }}132,{\rm{ }}213,{\rm{ }}231,{\rm{ }}312,{\rm{ }}321$.
So, Number of permutations of n things, taken r at a time, denoted by ${}^n{P_r} = \dfrac{{n!}}{{(n - r)!}}$
Combinations may be defined as the various ways in which objects from a set may be selected. For example: The different selections possible from the numbers 1, 2, 3 taking 2 at a time, are ${\rm{12, 23, and \,31}}$
So, Number of combinations possible from n group of items, taken r at a time, denoted by $^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$
Factorial of a number is equal to the multiplication or product of all the positive integers smaller than or equal to the number and factorial of 1 is always equals to 1 and also factorial of 0 is equals to 1. The factorial of a number is always positive it can never be negative and the factorial of a negative number is not defined.
Example of factorial:$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$