Question

The total number of words formed by the letters of the word 'PARABOLA', if only two A's should come together, is$A. 5040$
B. 3600$C. 4320$
D. 2400$$$$

Answer 1

We find the total number of words where only two A's should come together as $T-{{A}_{3}}-{{A}_{0}}$ where T is the total number of words that can be made from the letters of parabola, ${{A}_{3}}$ is the number of words where 3 A’s come together and ${{A}_{0}}$ is the number of words where no two A’s come together. $$$$

Complete step-by-step answer:
We see that in the word ‘PARABOLA' there are 8 letters with letter A repeated 3 times and the other 5 letters P,R,B,O,L which do not repeat. The total number of word we can from the letters of ‘PARABOLA' is the number of arrangements of 8 letters where one letter A repeats 3 times is
$T=\dfrac{8!}{3!}=4\times 5\times 6\times 7\times 8=6720$
We now find the number of ways the three A’s come together. Let us consider the three A’s a single letter and place them in close to each other in three consecutive places.

A & A & A & \\_ & \\_ & \\_ & \\_ & \\_ \\\ \end{matrix}$$ We can fill the rest empty5 places with non-repeating 5 letters P, R, B, O and L. So we have a total 5 plus 1 letter for AAA that is a total 6 letters which we can arrange in $6!$ ways. So the number of letters where three A’s come together is $${{A}_{3}}=6!=720$$ We now find the words where no two A’s come together. We can first fill non-repeating 5 letters P, R, B, O and L leaving an empty place on both sides of them as shown below. $$\begin{matrix} \\_ & P & \\_ & R & \\_ & B & \\_ & O & \\_ & L & \\_ \\\ \end{matrix}$$ We arrange the non-repeating 5 letters in $5!$ ways. We see in the above figure that there are 6 empty places where we can fill the 3 A’s in ${}^{6}{{C}_{3}}$ ways. We use rule of product and find the total number of words where no A’s come together as $${{A}_{0}}={}^{6}{{C}_{3}}\times 5!=20\times 120=2400$$ So the total number of words where two A’s come together is $$T-{{A}_{3}}-{{A}_{0}}=6720-720-2400=3600$$ **So, the correct answer is “Option B”.** **Note:** We have used here formulas for the selection of $r$ distinct places from $n$ distinct places as ${}^{n}{{C}_{r}}$, the arrangement of $n$ things where one things repeats itself $p$ times as $\dfrac{n!}{p!}$. We can alternatively solve directly by placing the non-repeating 5 letters P, R, B, O and L with empty 6 places beside them. We can arrange them in $5!$ ways. We treat two A’s as a single letter and one A as the other letter which we can place in the empty places in ${}^{6}{{C}_{2}}$ ways and then arrange them in $2!$ ways. So the total number of words will be $5!{{\times }^{6}}{{C}_{2}}\times 2!=3600$.