How many 4 letter words containing (G) can be formed using letters of DAUGHTER (repetition not allowed).

Hint: First find the number of selections possible in this case. Now you have selected words. So, now arrange them. Each case can be arranged in the same numbers of ways. So, multiply the numbers of cases with the arrangements. Now you get the total number of words possible. The ways of arranging in different things are given by: $n!$

Complete step-by-step answer:
Combinations: It is a mathematical technique that determines the number of possible arrangements in a collection of items where the order of selection does not matter in combinations you can select items in any order. Formula is given by
$^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
Given condition in the question are written as given follow:
4 letter words formed from DAUGHTER
Next given condition in the question can be written as given below:
1 letter selected will be G
As per given conditions; we need the remaining 3 letters to be selected.
As the letter, G is removed now we have 7 letters.
So, we need to select 3 letters from 7 letters.
By definition of combination, we get the number of ways:
$^{7}{{C}_{3}}$
By above we can say the number of way possible are
Number of ways= $^{7}{{C}_{3}}$
Now we have 4 different letters to be arranged. Here there are no repeated words or repeated selections. So, we have no repetition and all are different things.
We know number of arrangement n different things is:
$n!$
So, here we have 4 things which are all different.
By substituting n=4, we get the number of arrangements as:
4!
So, we get that each case has this number of arrangements. So, we must get the product of the number selections and number of arrangements, to get a number of 4 letters words. By using the above statement, we get the number of words = number of words = $^{7}{{C}_{3}}\times 4!$
By simplifying the above equation, we get it as:
Number of words = $\dfrac{7!}{4!\times 3!}\times 4!$
By cancelling the common terms, we get it as follows:
Number of words = $\dfrac{7!}{3!}=\dfrac{7\times 6\times 5\times 4\times 3!}{3!}$
By simplifying the above equation, we get the number of ways as:
Number of words = $7\times 6\times 5\times 4=840$
Therefore, there are 840 words possible with the given condition.

Note: Be careful while selecting. As G is already selected, we have only 7 possible from which we have to select. Generally, students forget and take 8. But you must take 7 possibilities only. But while making arrangements we must also consider the 8 to arrange the word’s letters. Do these steps carefully.