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# The number of four-letter words formed using the letters of the word INEFFECTIVE areA) 1200B) 1340C) 1422D) 1556

Last updated date: 23rd Jun 2024
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Hint: We will first count each letter and how many times it has occurred in the given word. We need to select four letters out of the available letters and form different combinations of them. Later we will add all the answers obtained in the process to conclude the final answer.

Complete step-by-step solution:
The given word is “INEFFECTIVE”.
We will start by counting the occurrence of each letter in the word.
The word “INEFFECTIVE” contains 3 Es, 2 Is, 2 Fs, and 1 N, C, T, V.
That implies we have 7 different kinds of letters out of which we have to select 4 letters.
Since there is a possibility that the letters might be similar so we will consider different possibilities.
The first possibility is very simple to consider. We will assume that all four letters are different.
In this case total, possible ways are given by permutations of 4 letters out of the given 7.
It is given by:
$\Rightarrow {}^7{P_4} = \dfrac{{7!}}{{\left( {7 - 4} \right)!}}$
Simplify the terms,
$\Rightarrow {}^7{P_4} = \dfrac{{7 \times 6 \times 5 \times 4 \times 3!}}{{3!}}$
Cancel out the common factors and multiply,
$\Rightarrow {}^7{P_4} = 840$ …………...….. (1)
The second possibility is that we will select two same letters and remaining both different letters.
The total possible ways to do so is given by:
$\Rightarrow {}^3{C_2} \times {}^6{C_2} \times {}^4{P_2} = \dfrac{{3!}}{{2!\left( {3 - 2} \right)!}} \times \dfrac{{6!}}{{2!\left( {6 - 2} \right)!}} \times \dfrac{{4!}}{{\left( {4 - 2} \right)!}}$
Simplify the terms,
$\Rightarrow {}^3{C_2} \times {}^6{C_2} \times {}^4{P_2} = \dfrac{{3 \times 2!}}{{2!1!}} \times \dfrac{{6 \times 5 \times 4!}}{{2 \times 1 \times 4!}} \times \dfrac{{4 \times 3 \times 2!}}{{2!}}$
Cancel out the common factors and multiply,
$\Rightarrow {}^3{C_2} \times {}^6{C_2} \times {}^4{P_2} = 540$..................….. (2)
The third possibility is the two same letters of each kind.
It can be given in the following ways:
$\Rightarrow {}^4{C_2} \times {}^3{C_2} = \dfrac{{4!}}{{2!\left( {4 - 2} \right)!}} \times \dfrac{{3!}}{{2!\left( {3 - 2} \right)!}}$
Simplify the terms,
$\Rightarrow {}^4{C_2} \times {}^3{C_2} = \dfrac{{4 \times 3 \times 2!}}{{2!2!}} \times \dfrac{{3 \times 2!}}{{2!1!}}$
Cancel out the common factors and multiply,
$\Rightarrow {}^4{C_2} \times {}^3{C_2} = 18$ ……………..….. (3)
The last possibility is three letters of the same kind and one of the different.
It can be given in the following ways:
$\Rightarrow {}^1{C_1} \times {}^6{C_1} \times {}^4{C_3} = \dfrac{{1!}}{{1!\left( {1 - 1} \right)!}} \times \dfrac{{6!}}{{1!\left( {6 - 1} \right)!}} \times \dfrac{{4!}}{{3!\left( {4 - 3} \right)!}}$
Simplify the terms,
$\Rightarrow {}^1{C_1} \times {}^6{C_1} \times {}^4{C_3} = \dfrac{{1!}}{{1!0!}} \times \dfrac{{6 \times 5!}}{{1!5!}} \times \dfrac{{4 \times 3!}}{{3!1!}}$
Cancel out the common factors and multiply,
$\Rightarrow {}^1{C_1} \times {}^6{C_1} \times {}^4{C_3} = 24$...............….. (4)
Thus, the total number of ways is given by adding equations (1) to (4),
$\Rightarrow 840 + 540 + 18 + 24$
Adding the terms we get the value as
$\Rightarrow 1422$

Hence, option (C) is correct.

Note: We considered the total number of different letters used in the given word. Later we considered all the possible combinations that can be formed for four-letter words. At some point, we used a combination while at some point we used permutation depending on the repetition required in the considered possibility.