Question

The number of arrangements that can be made out of the letters of the word SUCCESS so that all S do not come together is
A. 60
B. 120
C. 360
D. 420

Answer 1

We have given a word and we have to find the number of arrangements that can be made out of the word SUCCESS, so that all the S do not come together. Firstly, we see that there are $7$ letters in the word out of which there are $3$ S’ and $2$ C’s. We will calculate the total number of arrangements then we calculate the arrangements in which $3$ S’ are together. Required arrangement will be total arrangement minus number of arrangements in which $3$ S’ are together.

Complete step-by-step answer:
We have the word SUCCESS.
Total number of letters $= 7$
We have to find the number of arrangements in which all S’ do not come together.
Firstly, we have to calculate total arrangements.
Total arrangements of $n$ things in which $p$ are alike and $q$ are alike.
So, total arrangements$= \dfrac{{7!}}{{3! \times 2!}}$
Now, we calculate the arrangements in which $3$ S’ are together.
Here, we take $3$ S’ as one unit, so
Total letters $= 7 - 3 + 1$
$= 5$ letters
So, number of arrangements in which $3$ S’ are together $= \dfrac{{5!}}{{2!}} \times \dfrac{{3!}}{{3!}}$
$= \dfrac{{5!}}{{2!}}$
Now, required arrangements $=$ Total arrangements $-$ Number of arrangements in which $3$ S’ are together
$= \dfrac{{7!}}{{3! \times 2!}} - \dfrac{{5!}}{{2!}}$
$= \dfrac{{7 \times 6 \times 5!}}{{3! \times 2!}} - \dfrac{{5!}}{{2!}}$
$\dfrac{{5!}}{{2!}}$ take common from the above equation.
$= \dfrac{{5!}}{{2!}}\left[ {\dfrac{{7 \times 6}}{{3 \times 2 \times 1}} - 1} \right]$
Expanding the factorial of 5 and cancelling the common term in brackets.
$= \dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{2} \times \left[ {7 - 1} \right]$
$= \dfrac{{5 \times 4 \times 3 \times 2 \times 1 \times 6}}{2}$
$= 6 \times 5 \times 4 \times 3 \times 1$
$= 360$

So, number of arrangements in which $3$ S’ are not together $= 360$

Note: In permutation, the first place can be filled up in ‘$n$’ different ways as any one of the ‘$n$’ persons can be placed there. After filling up the first place in one of the ‘$n$’ ways. There are $\left( {n - 1} \right)$ different ways to fill up the second place. Similarly, the third place can be filled by $\left( {n - 2} \right)$ different ways.
Proceeding in this way, we see that whenever a pace if filled up by a new factor is introduced. The factor begins with n and goes on diminishing by unity.
Therefore, rth factor $= n - \left( {r - 1} \right)$
$= n - r + 1$
Therefore, number of ways filling up r places $= n\left( {n - 1} \right)\left( {n - 2} \right)......\left( {n - r + 1} \right)$
Therefore, ${}^n{p_r} = \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right).....\left( {n - r + 1} \right).....3.2.1}}{{\left( {n - r} \right).....3.2.1}}$
Therefore, ${}^n{p_r} = \dfrac{{\left| \\!{\underline {\, n \,}} \right. }}{{\left| \\!{\underline {\, n \,}} \right. - r}}$