Question

Prove that only 1422 different four letter words can be formed out of the letters of the word INEFFECTIVE.

Answer 1

Hint: We can observe that the word INEFFECTIVE has 4 different letters, namely N, C, T and V, i.e. 3 E, 2 I and 2 F. So we will take 4 different cases to form a four letter word from these letters. In the first case we will take all 4 different letters, in the second case we will take 2 same and 2 different letters, in the third case we will take 2 same and 2 same letters and the last case in which we will take 3 same and 1 different letters.
Complete step-by-step answer:
We have been given the word INEFFECTIVE.
We can observe that the word INEFFECTIVE has 4 different letters, namely N, C, T and V, i.e. 3E, 2I and 2F.
Now, we will take the first case in which all the letters of the four letter words are different.
Then the number of words would be as follows:

& ^{7}{{C}_{4}}\times 4!=\dfrac{7!}{\left( 7-4 \right)!4!}\times 4! \\\ & =\dfrac{7!}{\left( 7-4 \right)!}=\dfrac{7!}{3!}=\dfrac{7\times 6\times 5\times 4\times 3!}{3!} \\\ & =7\times 6\times 5\times 4=840 \\\ \end{aligned}$$ Since we have 7 different letters E, I, F, N, C, T and V so we will select and arrange 4 letters among these. Now, we will take the second case in which 2 letters are the same and the other two letters are different from the four letter words. Then the number of words would be as follows: $$\begin{aligned} & ^{3}{{C}_{1}}{{\times }^{6}}{{C}_{2}}\times \dfrac{4!}{2!}=3!\times \dfrac{6!}{\left( 6-2 \right)!\times 2!}\times \dfrac{4!}{2!} \\\ & =3!\times \dfrac{6!}{4!\times 2!}\times \dfrac{4!}{2!} \\\ & =3\times 2\times 1\times \dfrac{6\times 5\times 4\times 3\times 2!}{2!}\times \dfrac{4\times 3\times 2!}{2!} \\\ & =6\times 360\times 12=540 \\\ \end{aligned}$$ Since we have 3 sets in which 2 letters are the same, so we have to choose any one from them and 6 letters are there which we choose 2 letters and arrange all the 4 letters keeping in mind 2 letters are the same. So we will divide it by 2. Now we will take the third case in which 2 letters are of the same kind and the other 2 letters are of the same kind of the four letter words. We can also say that in this case, all letters of the word are the same. Then, the number of words would be as follows: $$\begin{aligned} & ^{3}{{C}_{2}}\times \dfrac{4!}{2!\times 2!}=\dfrac{3!}{\left( 3-2 \right)!2!}\times \dfrac{4!}{2!\times 2!} \\\ & =\dfrac{3!}{1\times 2!}\times \dfrac{4!}{2!\times 2!}=\dfrac{3\times 2!}{2!}\times \dfrac{4\times 3\times 2!}{2!\times 2\times 1} \\\ & =3\times \dfrac{12}{2}=3\times 6=18 \\\ \end{aligned}$$ Since we have 3 sets of letters in which at least 2 letters are of the same kind, so we will have to select 2 sets from them and arrange it. Also the 2 sets have the same letters so we will divide it by $$2!\times 2!$$. Once again we will take the fourth case in which there are 3 same letters and 1 different letter of the fourth letter word. Then the number of words would be as follows: $$\begin{aligned} & ^{1}{{C}_{1}}{{\times }^{6}}{{C}_{1}}\times \dfrac{4!}{3!}=1\times 6\times \dfrac{4\times 3!}{3!} \\\ & =1\times 6\times 4=24 \\\ \end{aligned}$$ Since there is one set in which all 3 are the same letters and 6 are different letters from which we select 1 letter and arrange all the four letters. Hence, total number of words $$=840+540+18+24=1422$$. Hence it is proved that only 1422 different words that can be formed from the letters of the word INEFFECTIVE. Note: Be careful while calculating the calculation part as there is a chance of multiplication error. Also remember that permutation means arrangement of things, the order of things is considered whereas combination means selection of things, the order of things has no importance.