Question

If letters of the word ‘GYRATION’ is permuted in all possible ways,
a) How many words begin with G
b) How many words start with G and end with T

& A.7!,6! \\\ & B.8!,7! \\\ & C.7!,7! \\\ & D.6!,7! \\\ \end{aligned}$$

Answer 1

Count the total number of letters present in the given word. In option (a) fix G at first place and arrange the remaining letters in the remaining places and in option (b) fix first and last place with G2T and proceed as option (a).

Complete step-by-step answer :
Before proceeding to solving the question, we should first understand what permutation of letters or anything means.
A permutation is a collection or a combination of objects from a set where the order or the arrangement of chosen object matter, that is, arrangement of objects in a definite order.
Suppose, there are n distinct items, out of which we have to choose r items. We can choose the items in following ways:
A) Selection when repetition of item is allowed.
B) Selection when repetition of items is not allowed.
With repetition of items allowed, at every step for n total items we have all n choices. For example:
PLAY = 4 letters.
All possible ways of permutation with repetition would be:

& \begin{matrix} {\bar{\uparrow }} & {\bar{\uparrow }} & {\bar{\uparrow }} \\\ \text{All 4} & \text{All 4} & \text{All 4} \\\ \text{All 4} & \text{letters} & \text{can be} \\\ \end{matrix}\text{ }\begin{matrix} \overline{{}} \\\ \text{All 4} \\\ \text{occupied} \\\ \end{matrix} \\\ & \Rightarrow 4\times 4\times 4\times 4={{4}^{4}}\Rightarrow {{4}^{n}} \\\ \end{aligned}$$ Without repetition of items, for example: PLAY $$\begin{aligned} & \begin{matrix} \underline{4} & \underline{3} & \underline{2} \\\ \uparrow & \uparrow & \uparrow \\\ \text{PL(A)Y} & \text{P(L)Y} & \text{(P)Y} \\\ \end{matrix}\text{ }\begin{matrix} \underline{1} \\\ \downarrow \\\ \text{Y} \\\ \end{matrix} \\\ & \Rightarrow \text{4}\times \text{3}\times \text{2}\times \text{1=4!} \\\ \end{aligned}$$ If we put A in first letter place then, only 3 letters would be remaining, in next step only 2 letters and in the last step only 1 letter would be available. Now, coming to the question word GYRATION. This word contains a total 8 letters or n = 8. Option (a): if we fix the first place as G. $$\text{ }\begin{matrix} \underline{G} \\\ \uparrow \\\ \text{Fixed} \\\ \end{matrix}\overline{7}\text{ }\overline{6}\text{ }\overline{5}\text{ }\overline{4}\text{ }\overline{3}\text{ }\overline{2}\text{ }\overline{1}$$ Now, we are left with only 7 letters, 7 letters are available for next step, then 6 letters, then 5 letters and this goes on till the last step where only 1 letter would be available. So, $$\begin{aligned} & \text{P = Permutation of required condition} \\\ & \Rightarrow \text{7}\times \text{6}\times \text{5}\times \text{4}\times \text{3}\times \text{2}\times \text{1} \\\ & \Rightarrow \text{7!} \\\ \end{aligned}$$ Option (b): now if we fix both first and last places, the remaining 6 letters can be arranged. $$\begin{matrix} \underline{G} \\\ \uparrow \\\ \text{Fixed} \\\ \end{matrix}\text{ }\begin{matrix} \overline{\uparrow } \\\ 6 \\\ \end{matrix}\text{ }\begin{matrix} \overline{\uparrow } \\\ 5 \\\ \end{matrix}\text{ }\begin{matrix} \overline{\uparrow } \\\ 4 \\\ \end{matrix}\text{ }\begin{matrix} \overline{\uparrow } \\\ 3 \\\ \end{matrix}\text{ }\begin{matrix} \overline{\uparrow } \\\ 2 \\\ \end{matrix}\text{ }\begin{matrix} \overline{\uparrow } \\\ 1 \\\ \end{matrix}\text{ }\begin{matrix} \underline{T} \\\ \uparrow \\\ \text{Fixed} \\\ \end{matrix}$$ So, $$\begin{aligned} & \text{P=6}\times \text{5}\times \text{4}\times \text{3}\times \text{2}\times \text{1} \\\ & \Rightarrow \text{6!} \\\ \end{aligned}$$ Therefore, option A is correct. **Note** : We did not use the permutation condition with repetition allowed because it is not mentioned here in the question. If not mentioned anything, proceed with permutation without repetition allowed. Some students tend to apply the concept of combination here, but they must note that here we have been asked to find possible arrangements and not ways of selecting something. So, do not use combination formulas and solve the question.