Question

In how many ways can the letter of the word SUCCESS be arranged so that
(1)The two${{C}^{'}}s$are together but no two${{S}^{'}}s$are together.
(2)No two${{C}^{'}}s$and no two${{S}^{'}}s$are together.

Answer 1

Hint: First see what is asked. Then take a bundle of${{S}^{'}}s$and${{C}^{'}}s$. Use $=m!\times {}^{m+1}{{C}_{n}}\times n!$and you will get the answer.

Complete step-by-step answer:

So we are given the word SUCCESS,

So we want to find the given above.

So for that first we should find in how many ways the word can be arranged with repeating,
So the formula for permutations with repeated elements is as follows, when$k$out of$n$elements are indistinguishable. For example: $k$copies of the same book, the number of different permutations is$\dfrac{n!}{k!}$.

So the total number of arrangement of the word SUCCESS can be formed $=$$\dfrac{7!}{3!2!}$=$\dfrac{7\times 6\times 5\times 4\times 3!}{3!\times 2\times 1}=420$ways

So SUCCESS and be arranged in$420$ways.



Now taking,

 (1) The two${{C}^{'}}s$are together but no two${{S}^{'}}s$are together.

Now in word SUCCESS, there are$3$${{S}^{'}}s$and$2$${{C}^{'}}s$,

Now if we take

So let us take two ${{C}^{'}}s$one bundle, So the remaining letters are $5$,

That is we have total $6$, So Now let us choose seat for two ${{C}^{'}}s$which is${}^{6}{{C}_{1}}$.

Thus we now have $5$blanks left,

So let us choose a seat for the letter$U$and$E$which we get as${}^{5}{{C}_{2}}$.

So then the arrangement of$U$and$E$will be $2!$,

So three${{S}^{'}}s$should not be together.

Now we have to use gap method,

Gap Method: The number of permutations when no two given objects occur together. In order to find the number of permutations when no two given objects occur together.

(a) First of all, put the$m$objects for which there is no restriction, in a line. These$m$ objects can be arranged in$m!$ ways.

(b) Then count the number of gaps between every two$m$objects for which there is no restriction, including the end positions. Number of such gaps will be$(m+1)$.

(c) If $m$is the number of objects for which there is no restriction and$n$is the number of objects, two of which are not allowed to occur together, then the required number of ways $=m!\times {}^{m+1}{{C}_{n}}\times n!$

Now the number of blanks filled are $3$.

And the number of gaps formed $=(3+1)=4$

So now taking the gaps for$S$to fill$={}^{4}{{C}_{3}}$

So for the two${{C}^{'}}s$are together but no two${{S}^{'}}s$are together\[={}^{6}{{C}_{1}}\times {}^{5}{{C}_{2}}\times 2!\times 3\times {}^{4}{{C}_{3}}=\dfrac{6!}{1!(6-1)!}\times \dfrac{5!}{2!(5-2)!}\times 2\times \dfrac{4!}{3!(4-3)!}\]

So simplifying above in simple manner we get,

\[=6\times 10\times 2\times 4=480\]

So for the two${{C}^{'}}s$are together but no two${{S}^{'}}s$are together the word SUCCESS can be arranged in the $480$ways.


(2) No two${{C}^{'}}s$and no two${{S}^{'}}s$are together.

Now we first choose the seats for$U$and$E$, there are two letters$U$and$E$so it becomes${}^{7}{{C}_{2}}$,

So now arrangements of$U$and$E$$=2!$

So using gap method so number of gaps formed$=(2+1)=3$

Now choosing gaps for${{C}^{'}}s$which is${}^{3}{{C}_{2}}$,

For ${{S}^{'}}s$,

So now we get the number of gaps formed as$=(2+2+1)=5$

Now choosing gaps for ${{S}^{'}}s$which we get as${}^{5}{{C}_{3}}$,

So for No two${{C}^{'}}s$and no two${{S}^{'}}s$are together$={}^{7}{{C}_{2}}\times 2!\times {}^{3}{{C}_{2}}\times {}^{5}{{C}_{3}}=\dfrac{7!}{2!(7-2)!}\times 2\times \dfrac{3!}{2!(3-2)!}\times \dfrac{5!}{3!(5-3)!}$

So simplifying above we get,

$=21\times 2\times 3\times 10=1260$

So for the No two${{C}^{'}}s$and no two${{S}^{'}}s$are together the word SUCCESS can be arranged in the $1260$ways.

Note: So read the question properly what is asked. So you should know how we can take a set of three. The concept regarding repetitive and no repetitive should be clear. You should be familiar with the formula$=m!\times {}^{m+1}{{C}_{n}}\times n!$and also the gap method. Also you should be clear with the fundamental theorem and how to compute it.