Question

How many words can be formed from the letters of the word COURTESY, whose first letter is C and last letter is Y.
$$\begin{gathered} \text{A}\text{. 6!} \\ \text{B}\text{. 8!} \\ \text{C}\text{. 2}\left(6!\right) \\ \text{D}\text{. 2}\left(7!\right) \end{gathered}$$

Answer 1

Hint: We have given word COURTESY and condition that first letter should be C and last letter should be Y that means we have 8 letters in the word in which 2 is bounded that is one at first position and other at last position so we have to find possibility for 6 letters on six places for which we will take help of permutation and combination.

Complete step-by-step answer:
We have given word COURTESY
$\text{C,}\dots \text{,}\dots \text{,}\dots ,\dots ,\dots ,\dots ,\text{Y}$
Here we have fixed the first letter as C and the last letter as Y as given in the question.
Now we have six letters O,U,R,T,S,E and we have six spaces in which we have to arrange these six letters.
So now, for the first space we have six choices because we have six letters and we can put any one of them so no ways will be = 6.
For second space we have remaining five letters in which we can select any one of them so number of ways will be = 5
Now for third space we have four letters remaining in which we can choose any one of them so number of ways will be = 4
And for fourth space we have remaining letters is 3 so we have only three choices so number of ways will be = 3
Now for fifth space we have remaining only two letters so we have to choose from two hence number of ways will be = 2
And at last for sixth place we have one and only one letter remaining and so we have only one option hence the number of ways will be = 1.
And we have to fill all the spaces so from product rule of permutation and combination
Total number of words formed will be = $6\times 5\times 4\times 3\times 2\times 1=6!$
Hence option A is the correct option.

Note: Whenever we get this type of question the basic way of solving is explained but we know directly that if we have n objects and we have to fill ‘n’ spaces then the total number of ways will be $n!$. and here product rule is used so we should have knowledge about this rule. The Product Rule: If there are n(A) ways to do A and n(B) ways to do B, then the number of ways to do A and B is $n(A)\times n(B)$.