Question

how many arrangements of the word ARRANGE can be made
I.If the two R’s are not allowed to come together, let it be k?
II.If neither the two R’s nor the two A’s are allowed to come together, let it be m.
Find the sum of digits of m+k ?

Answer 1

Hint: The arrangement of words can be solved with the help of the Factorials. Factorial is the product of all values until 1, for suppose, the value of the factorial of the 2 is taken as\[2 \times 1 = 2\]. Arrangement is the concept that can be defined as the different ways that a word can be placed or arranged.

Complete step-by-step answer:
The word is ARRANGE,
Number of words in the word ARRANGE is \[X = 7\].
Number of R’s in the word AARANGE is \[y = 2\].
Number of A’s in the word ARRANGE is \[z = 2\].
Number of repeated words in the word ARRANGE is 2 (A and R).
i)
We know the equation to find that two R’s are not allowed to come together is given as,
\[k = \dfrac{{X!}}{{y!.z!}} - \dfrac{{6!}}{{y!}}\]
Here, k is number of possible arrangement.
On substituting the values in the equation, then we get,
\[\begin{array}{c}
k = \dfrac{{7!}}{{2!.2!}} - \dfrac{{6!}}{{2!}}\\
k = \dfrac{{5040}}{4} - \dfrac{{720}}{2}\\
k = 900
\end{array}\]
Therefore, the two R’s are not allowed to come together in word ARRANGE or k is 900.
ii)
We know the equation to find neither the two R’s nor the two A’s are allowed to come together is,
\[m = \dfrac{{X!}}{{y!.z!}} - 5!\]
On substituting the values in the equation, then we get,
\[\begin{array}{c}
m = \dfrac{{7!}}{{2!.2!}} - 5!\\
m = \dfrac{{5040}}{4} - 120\\
m = 1140
\end{array}\]
Therefore, the value of m is 1140.
Now, we will find the sum of the k and m, we obtain,
\[\begin{array}{c}
k + m = 900 + 1140\\
 = 2040
\end{array}\]

Note: Here, we should be careful while framing the equation, the total words is placed in the numerator and number of words that number of times repeated is placed in the denominator. For the first part, the R is repeated two times, so we have to assume the two R’s as one which give 6 and number of times. And coming to the second part the A and R are taken as one and assuming as the 5 letters so we took \[5!\] in it.