Question

How many permutations for a 7 characters in length string, which contains all following letters R, X, S, Y, T, Z, U, has either the string 'RXS' or string 'ZU' in the 7 characters string?

Answer 1

Answer:

Following are the solution to this question:

Explanation:

In the following forms, RXS can appear:  

$R X S \_ \_ \_ \_$ it may look like that really, $4 \times 3 \times 2 \times 1$ forms = 24 may construct the remainder of its letters.  

$\_ R X S \_ \_ \_$ it may look like that really,  $4 \times 3 \times 2 \times 1$ forms = 24 may construct the remainder of its letters.  

$\_ \_ R X S \_ \_$ it may look like that really, $4 \times 3 \times 2 \times 1$ forms = 24 may construct the remainder of its letters.  

$\_ \_ \_ R X S \_$ it may look like that really,  $4 \times 3 \times 2 \times 1$ forms = 24 may construct the remainder of its letters.  

$\_ \_ \_ \_ R X S$ it may look like that really,  $4 \times 3 \times 2 \times 1$ forms = 24 may construct the remainder of its letters.  

And we'll have a total of $24 \times 5 = 120$ permutations with both the string RXS.

In the following forms, UZ can appear:

$U Z \_ \_ \_ \_\ _$  They can organize your remaining 5 characters through 5 categories! Procedures $= 5 \times 4 \times 3 \times 2 \times 1 = 120$

$\_ UZ \_ \_ \_ \_$ They can organize your remaining 5 characters through 5 categories! Procedures $= 5 \times 4 \times 3 \times 2 \times 1 = 120$

$\_ \_ U Z \_ \_ \_$They can organize your remaining 5 characters through 5 categories! Procedures  $= 5 \times 4 \times 3 \times 2 \times 1 = 120$

$\_ \_ \_ U Z \_ \_$They can organize your remaining 5 characters through 5 categories! Procedures  $= 5 \times 4 \times 3 \times 2 \times 1 = 120$

$\_ \_ \_ \_ U Z \_$ They can organize your remaining 5 characters through 5 categories! Procedures  $= 5 \times 4 \times 3 \times 2 \times 1 = 120$

$\_ \_ \_ \_ \_ U Z$ They can organize your remaining 5 characters through 5 categories! Procedures  $= 5 \times 4 \times 3 \times 2 \times 1 = 120$

There may be $120 \times 6 = 720$ ways of complete permutations.