How many different four-letter passwords can be formed from the letters upper a comma upper b comma upper c comma upper d comma upper e comma upper f comma and upper g if no repetition of letters is allowed?

Answer with explanation:

Number of Letters from which we have to make four-letter passwords

   ={A,B,C,D,E,F,G}=7

Since repetition of digit is not allowed.In Permutation order of arrangement is Important,while in combination order of arrangement is not Important.

Out of seven letters we have to select four letters, keeping in mind the order of Alphabets.So, we will use the concept of Permutation.

Number of ways of arrangement,that is to make four letter password from 7 Alphabets

So, total number of ways of making password from 7 alphabets=840 ways

Second Method⇒

Since repetition of alphabets is not allowed, first Alphabet can be chosen in 7 ways , second alphabet can be chosen in 6 ways, third alphabet can be chosen in 5 ways, fourth alphabet can be chosen from four alphabets.

So, total number of ways of making four letter password

       =7×6×5×4

    = 840 ways

the length of o'p' is .

step-by-step explanation:

it is given that the quadrilateral m'n'o'p' is the image of quadrilateral mnop after dilation.

the center of dilation is origin and the scale factor is 1.5.

in dilation the image and preimge are similar figures. it means the sides of image and preimage are proportional.

the scale factor is defined as

the scale factor is 1.5.

multiply both sides by op.

note: if the value of op is given, then multiply the value of op with 1.5 to get the length of side o'p'.

therefore length of o'p' is .