Question

The number of ways in which four letters of the word MATHEMATICS can be arranged is given by:
A. 136
B. 192
C. 1680
D. 2454

Answer 1

First we will find the sum of the three cases, where the case 1 is with two alike and other two alike letters, case 2 is two alike, two different letters and case 3 is with all are different letters. Then simplify using the formula to calculate 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 to find the required answer.

Complete step by step answer:

We are given the word “MATHEMATICS”.

As we can see that there are some letters in the given word, which are getting repeated.
So, while selecting the letters for arrangement we will consider all the cases.
First, since we have 11 letters in the given word "Mathematics" in which there are 2 M's, 2 T's, 2 A's, and other letters single.
Case 1:
Two alike and other two alike letters
In this case, we will select the two letters, which are alike.
Since we have three choices M, T, A, find the number of ways so that we have to select two letters out of three, we get
$\Rightarrow {}^3{C_2}$
Using the formula to calculate 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 in the above expression, we get

$$\Rightarrow \dfrac{{3!}}{{2!\left( {3 - 2} \right)!}} \\\ \Rightarrow \dfrac{{3 \times 2!}}{{2!1!}} \\\ \Rightarrow 3{\text{ ......eq.(1)}} \\\$$

Thus, there are 3 ways.
Case 2:
One alike and two different letters
In this case we will select 1 alike letter and other 2 different letters, so finding the number of ways for it, we get
$\Rightarrow {}^3{C_1} \times {}^7{C_2}$
Using the formula to calculate 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, we get

$$\Rightarrow \dfrac{{3!}}{{1!\left( {3 - 1} \right)!}} \times \dfrac{{7!}}{{2!\left( {7 - 2} \right)!}} \\\ \Rightarrow \dfrac{{3!}}{{1!2!}} \times \dfrac{{7!}}{{2!5!}} \\\ \Rightarrow \dfrac{{3 \times 2!}}{{1!2!}} \times \dfrac{{7 \times 6 \times 5!}}{{2!5!}} \\\ \Rightarrow 3 \times 21 \\\ \Rightarrow 63{\text{ .......eq.(2)}} \\\$$

Thus, there are 63 ways.
Case 3:
All are different letters
Computing the number of ways for 8 different letters, we get
$\Rightarrow {}^8{C_4}$
Using the formula to calculate 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, we get

$$\Rightarrow \dfrac{{8!}}{{4!\left( {8 - 4} \right)!}} \\\ \Rightarrow \dfrac{{8 \times 7 \times 6 \times 5 \times 4!}}{{4!4!}} \\\ \Rightarrow \dfrac{{8 \times 7 \times 6 \times 5}}{{24}} \\\ \Rightarrow 28 \times 5 \\\ \Rightarrow 140{\text{ ......eq.(3)}} \\\$$

Thus, there are 140 ways.
Finding the arrangement of each the cases and multiplying the equation (1) by ${}^4{C_2}$, (2) by ${}^4{C_2}$and (3) by $4!$, we get

$$\Rightarrow 3 \times \dfrac{{4!}}{{2!\left( {4 - 2} \right)!}} + 63 \times \dfrac{{4!}}{{2!}} + 140 \times 4! \\\ \Rightarrow 3 \times \dfrac{{4!}}{{2!2!}} + 63 \times 12 + 140 \times 24 \\\ \Rightarrow 3 \times 6 + 756 + 1680 \\\ \Rightarrow 18 + 756 + 1680 \\\ \Rightarrow 2454{\text{ ways}} \\\$$

Thus, option D is correct.

Note: In solving these types of questions, generally students get confused between combination and permutation. If you have to use combination ${}^n{C_r}$ and when you have to arrange use the formula of permutation ${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$, it is very tricky to use. Some students forget to consider all possibilities or else might get the wrong answer, because if some case is missed then it will result in the wrong answer.