Question

No of seven letter words that can be formed by using the letters of the word SUCCESS so that:
(a)Two C are together but no two S are together be k.
(b)No two C and no two S are together be m
Find m/k?

Answer 1

Hint: this can be solved using number of permutations of “n” different things taken all at a time is \[\left| \!{\underline {\,
  n \,}} \right. \]and number of permutations of “n” things taken all at a time of which “p” are alike is \[\dfrac{\left| \!{\underline {\,
  n \,}} \right. }{\left| \!{\underline {\,
  p \,}} \right. }\]. So by using these two formulas we can approach the answer.

Complete step-by-step answer:
(a)Consider CC as single object U, CC, E can be arranged in \[\left| \!{\underline {\,
 3 \,}} \right. \]ways
\[\times U\times CC\times E\times \]
Now 3 S has to be placed in the four available places
Hence required number of ways is \[=\left| \!{\underline {\,
  3 \,}} \right. \times {}^{4\mathop{C}_{3}}\]
\[=24\]
So, total no of ways in which two C are together but no two S are together be 24
Hence value of k is 24
(b)Let us find the words in which no two S are together
\[\times U\times C\times C\times E\times \]
The remaining letters U, C, C, E can be arranged in \[\dfrac{\left| \!{\underline {\,
  4 \,}} \right. }{\left| \!{\underline {\,
  2 \,}} \right. }=12\]ways
Now there are 5 five available places for three SSS
Hence total number of ways in which no two S are together is \[=12\times {}^{5\mathop{C}_{3}}\]
\[=120\]
Hence no of words having CC separated and SSS separated \[=120-24\]
\[=96\]
Hence total number of ways in which no two C and no two S are together be 96
Now we have the values of k=24,m=96
Then the value of \[\dfrac{m}{k}\]is
\[\dfrac{96}{24}=4\]

Note: In the above problem we have used permutations because it is only involved in arrangement of things whereas combinations are used for selection of things. Each of different things can be made by taking some or all of a given number of things or objects at a time is called permutation