Question

Find the number of combinations in the letters of the word STATISTICS taken $$4$$ at a time.

Answer 1

Hint: There are ten letters in the word out of which five are different. Since we are choosing four letters, we have to split it into different cases according to the letter repetition. For each case, we can calculate the number of combinations and adding we get the total.

Formula used: We can choose $$r$$ items from $n$ items in ${}^n{C}_r$ ways, where ${}^n{C}_r={\displaystyle \frac{n!}{(n-r)!r!}}$.

Complete step-by-step answer:
Given, the word “STATISTICS”.
We have to find the number of combinations in the letters of the word STATISTICS taken $$4$$ at a time.
The four letters chosen may be of four types.
The word contains ten letters in which five are different.
There are $3-S,3-T,2-I,1-A,1-C$
We can check each case separately.
Case (i): $3$ letters are the same and $1$ is different.
There are $2$ letters ($S,T$) which are repeated thrice.
So, in this case we have ${}^2{C}_1$ ways to choose the letter which is three times.
And the remaining letter can be either one among the four letters.
So, the total number of ways the selection can be made is $${}^2{C}_1\times 4=2\times 4=8$$
Therefore case (i) has $8$ ways.
Case (ii): Among the four two are the same.
That is, this case consists of two pairs of same letter combinations.
We can see two letters are the same and can be chosen from $S,T\mathrm{\&}I$.
So, there are ${}^3{C}_1$ ways.
${}^3{C}_1={\displaystyle \frac{3!}{(3-1)!1!}}={\displaystyle \frac{3\times 2\times 1}{2\times 1\times 1}}=3$
Number of ways in case (i) is $3$.
Case (iii): Two are the same and remaining two are different.
Then the same two can be chosen from $S,T\mathrm{\&}I$ and the different two can be chosen from the remaining four.
So, the same two can be chosen in ${}^3{C}_1$ ways and different two can be chosen in ${}^4{C}_2$ ways.
${}^3{C}_1={\displaystyle \frac{3!}{(3-1)!1!}}={\displaystyle \frac{3\times 2\times 1}{2\times 1\times 1}}=3$
${}^4{C}_2={\displaystyle \frac{4!}{(4-2)!2!}}={\displaystyle \frac{4\times 3\times 2\times 1}{2\times 1\times 2\times 1}}={\displaystyle \frac{12}{2}}=6$
So, the number of ways in case (iii) is $3\times 6=18$.
Case (iv): All are different
There are four positions and five different letters. So, the number of ways is ${}^5{C}_4$.
${}^5{C}_4={\displaystyle \frac{5!}{(5-4)!4!}}={\displaystyle \frac{5\times 4\times 3\times 2\times 1}{1\times 4\times 3\times 2\times 1}}=5$
So, the number of ways in case (iv) is $5$.
Therefore the total number of combinations in the letters of the word STATISTICS taken $$4$$ at a time is $8+3+18+5=34$.

So, the answer is $34$.

Note: If in case the letters given in the word are all different we can easily calculate the answer by a single case. Here the subcases occurred because of the repetition. And also if we were asked to choose the four letters without repetition, we have only the last case.