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# The number of permutations of the letters of the word INDEPENDENCE taken 4 at a time so that all the 4 are different isA.24B.120C.240D.360

Last updated date: 20th Sep 2024
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Hint: Here, we will use the selection that all the 4 are different using the combinations
${}^n{C_r} = \dfrac{{\left. {\underline {\, n \,}}\! \right| }}{{\left. {\underline {\, r \,}}\! \right| \cdot \left. {\underline {\, {n - r} \,}}\! \right| }}$, where $n$ is the number of items, and $r$ represents the number of items being chosen. Then simplify to find the required value.

We are given that the word is INDEPENDENCE.
We have 1 I’s, 3 N’s, 2 D’s, 4 E’s, 1 P’s and 1 C’s from the given word.
So, we know that there are four different letters in the given word, so we have $\left. {\underline {\, 4 \,}}\! \right|$.
We will find the selection that all the 4 are different using the combinations
${}^n{C_r} = \dfrac{{\left. {\underline {\, n \,}}\! \right| }}{{\left. {\underline {\, r \,}}\! \right| \cdot \left. {\underline {\, {n - r} \,}}\! \right| }}$, where $n$ is the number of items, and $r$ represents the number of items being chosen.
Here, there are 8 teams and each game is played by 2 teams, so we have
$n = 6$
$r = 4$
Substituting these values of $n$ and $r$ in ${}^n{C_r} = \dfrac{{\left. {\underline {\, n \,}}\! \right| }}{{\left. {\underline {\, r \,}}\! \right| \cdot \left. {\underline {\, {n - r} \,}}\! \right| }}$, we get
${}^6{C_4} = \dfrac{{\left. {\underline {\, 6 \,}}\! \right| }}{{\left. {\underline {\, 4 \,}}\! \right| \cdot \left. {\underline {\, {6 - 4} \,}}\! \right| }} \\ = \dfrac{{\left. {\underline {\, 6 \,}}\! \right| }}{{\left. {\underline {\, 4 \,}}\! \right| \cdot \left. {\underline {\, 2 \,}}\! \right| }} \\ = \dfrac{{6 \cdot 5 \cdot \left. {\underline {\, 4 \,}}\! \right| }}{{\left. {\underline {\, 4 \,}}\! \right| \cdot \left. {\underline {\, 2 \,}}\! \right| }} \\ = \dfrac{{6 \cdot 5}}{2} \\ = 15 \\$
Then we will multiply the above value with $\left. {\underline {\, 4 \,}}\! \right|$ to find the total number of ways, we get
$= 15 \times \left. {\underline {\, 4 \,}}\! \right| \\ = 15 \times 4 \times 3 \times 2 \times 1 \\ = 360 \\$
Therefore, the total number of games played is 360.
Hence, the option D is correct.

Note: We will divide the obtained probabilities of the total ways for selecting by the ways such that no constraints on quantity, so not multiply them. Here, students must take care while simplifying the conditions given in the question into the combinations. Some students use the formula of permutation, ${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$ instead of combinations is ${}^n{C_r} = \dfrac{{\left. {\underline {\, n \,}}\! \right| }}{{\left. {\underline {\, r \,}}\! \right| \cdot \left. {\underline {\, {n - r} \,}}\! \right| }}$, where $n$ is the number of items, and $r$ represents the number of items being chosen, which is wrong.