Question

arrangements with restrictions eg 1. how many arrangements of the letters of the word remand are possible if: a) there are no restrictions? solution: ( _6p_6 = 720 ) or ( 6! ) b) they begin with re? solution: re---- ( = _4p_4 = 24 ) or ( 4! ) c) they do not begin with re? solution: total - (b) ( = 6! - 4! = 696 )

Explanation:

Part (a)

Step1: Identify permutation type

We need to find the number of permutations of 6 distinct letters (R, E, M, A, N, D). The formula for permutations of \( n \) distinct objects is \( _nP_n=\frac{n!}{(n - n)!}=n! \).

Step2: Calculate \( 6! \)

\( 6! = 6\times5\times4\times3\times2\times1=720 \).

Step1: Fix "RE" at the start

If the arrangement begins with "RE", we consider "RE" as a single unit. Then we need to arrange the remaining 4 letters (M, A, N, D). The number of permutations of \( n \) distinct objects is \( _nP_n = n! \), here \( n = 4 \).

Step2: Calculate \( 4! \)

\( 4! = 4\times3\times2\times1 = 24 \).

Step1: Find total permutations

From part (a), total permutations of 6 letters is \( 6! = 720 \).

Step2: Find permutations starting with "RE"

From part (b), permutations starting with "RE" is \( 4! = 24 \).

Step3: Subtract to get the result

Subtract the number of permutations starting with "RE" from total permutations: \( 6!-4! = 720 - 24=696 \).

Answer:

720

Part (b)