Answer:
Following are the solution to this question:
Explanation:
In the following forms, RXS can appear:
it may look like that really,
forms = 24 may construct the remainder of its letters.
it may look like that really,
forms = 24 may construct the remainder of its letters.
it may look like that really,
forms = 24 may construct the remainder of its letters.
it may look like that really,
forms = 24 may construct the remainder of its letters.
it may look like that really,
forms = 24 may construct the remainder of its letters.
And we'll have a total of
permutations with both the string RXS.
In the following forms, UZ can appear:
They can organize your remaining 5 characters through 5 categories! Procedures ![= 5 \times 4 \times 3 \times 2 \times 1 = 120](https://tex.z-dn.net/?f=%3D%205%20%5Ctimes%204%20%5Ctimes%20%203%20%5Ctimes%202%20%5Ctimes%201%20%3D%20120)
They can organize your remaining 5 characters through 5 categories! Procedures ![= 5 \times 4 \times 3 \times 2 \times 1 = 120](https://tex.z-dn.net/?f=%3D%205%20%5Ctimes%204%20%5Ctimes%20%203%20%5Ctimes%202%20%5Ctimes%201%20%3D%20120)
They can organize your remaining 5 characters through 5 categories! Procedures ![= 5 \times 4 \times 3 \times 2 \times 1 = 120](https://tex.z-dn.net/?f=%3D%205%20%5Ctimes%204%20%5Ctimes%20%203%20%5Ctimes%202%20%5Ctimes%201%20%3D%20120)
They can organize your remaining 5 characters through 5 categories! Procedures ![= 5 \times 4 \times 3 \times 2 \times 1 = 120](https://tex.z-dn.net/?f=%3D%205%20%5Ctimes%204%20%5Ctimes%20%203%20%5Ctimes%202%20%5Ctimes%201%20%3D%20120)
They can organize your remaining 5 characters through 5 categories! Procedures ![= 5 \times 4 \times 3 \times 2 \times 1 = 120](https://tex.z-dn.net/?f=%3D%205%20%5Ctimes%204%20%5Ctimes%20%203%20%5Ctimes%202%20%5Ctimes%201%20%3D%20120)
They can organize your remaining 5 characters through 5 categories! Procedures ![= 5 \times 4 \times 3 \times 2 \times 1 = 120](https://tex.z-dn.net/?f=%3D%205%20%5Ctimes%204%20%5Ctimes%20%203%20%5Ctimes%202%20%5Ctimes%201%20%3D%20120)
There may be
ways of complete permutations.