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

Hint: 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.
\[\begin{matrix}
   A & A & A & \_ & \_ & \_ & \_ & \_ \\
\end{matrix}\]
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 $5!{{\times }^{6}}{{C}_{2}}\times 2!=3600$.