Question

Consider the word ${}^{\prime \prime }SUCCES{S}^{\prime \prime }$, then the total number of possible words

• A. when all the letters are taken at a time is $420$
• B. when All $S^{\prime }s$ are not together is $360$
• C. when no two $S^{\prime }s$ are together is $180$
• D. taking $4$ letters at a time is $92$

Answer 1

Answer: (A), (B)

when All $S^{\prime }s$ are not together is $360$

We have, $2C^{\prime }s,1E,3S^{\prime }s,1U$
Total permutations when all the letters are used $=\frac{7!}{3!2!}=420$

When All $S^{\prime }s$ are not together.
Total ways when All $S^{\prime }s$ are together $=\frac{5!}{2!}=60$
Required number of ways $=420-60=360$

When no two $S^{\prime }s$ are together.
Arrange other words in $4!=24$ ways.
Now the $S^{\prime }s$ can be arrnged between the gaps in ${}^5{C}_3=10$ ways
Required number of ways $=24\times 10=240$

Possible words taking $4$ letters at a time.

Case $1:$ When all the four letters are distinct:-
${}^4{P}_4=4!=24$

Case $2$: When only two letters are repeated:-
the repeated letter can be selected from $C,S$ i.e. ${}^2{C}_1$ and the remaining two letters can be selected from $3$ i.e. ${}^3{C}_2.$ After selecting these $4$ letters we have to arrange them while keep in mind that $2$ of them are same.
Number of ways
$=({}^2{C}_1\times {}^3{C}_2)\frac{4!}{2!}$
$=72$

Case $3$: When one is repeated twice and another one is repeated twice:-
the repeating letter can be selected as $C,S$ i.e. ${}^2{C}_2$
Number of ways
$={}^2{C}_2\times \frac{4!}{2!2!}$
$=6$

Case $4$: When one is repeated thrice and other is once:-
Number of ways:
$({}^1{C}_1\times {}^3{C}_1)\frac{4!}{3!}$
$=1\times 3\times 4$
$=12$
Total words $=24+72+6+12=114$