Question

In: Statistics and Probability

Customer account "numbers" for a certain company consist of 2 letters followed by 3 numbers. Step...

Customer account "numbers" for a certain company consist of 2 letters followed by 3 numbers.

Step 1 of 2 : How many different account numbers are possible if repetitions of letters and digits are allowed?

Solutions

Expert Solution

Here we have 5 blanks to fill.

Among the 5 blanks , the first two blanks should contain letters and the rest 3 should contain numbers

Now there are 26 letters in english alphabets and 10 numbers(0,1,2,3,4,5,6,7,8,9).

now first we consider the the first two places .

For the first place we can place any letter out of 26. Hence for the first place we have 26 choices.

Now for whatever choice we make for the first place it doesn't affect the choice of the second place and the 3rd, 4th,5th since we are said that repititions are allowed.

Suppose we choose A for the first blank then for the second blank we can choose any letter from A to Z that is total of 26 letters.

Therefore for a particular choice of letter in first place we can have 26 choices for the second place.Hence for a sigle selecetd letter in first place we can have 26 combination ,i.e suppose we choose A for the first place AA_ _ _ , AB_ _ _ , AC _ _ _ ,..........,AZ _ _ _ hence there are 26 possible combination for A selected in first place. Similarly for B in first place we can have BA_ _ _ ,BB _ _ _,....... BZ_ _ _ i.e again 26. also for C in first place we have 26 choices.

Therefore for all 26 possible choices of letters for the first place we have 26*26 possible combination.

Now for for the digits

Let us consider the combination AA for the first two place than for the third place we can have any of the 10 digits

i.e AA0_ _,AA1_ _,.......,AA9_ _ _. i.e total of 10 possibilities.similarly for AB0_ _ ,AB1_ _ _, .....,AB9_ _ _.10 possibilities

Hence for a selected pair of letters from all 26*26 possible combination we have 10 choices for the first place of the digit(3rd place all together). Hence total number of combination that are possible 26*26*10

Now for the 2nd place(i.e 4 th place all together) of the digits we also have 10 possible coices. eg- AB00_,AB01_,....,AB09_. hence 10 possible combination,

hence in the same way as above for a selected triplet from 26*26*10 combinations we have 10 options for the 4th place. Hence total possible combination 26*26*10*10.

Hence in the same way we can say that for a selected pair from 26*26*10*10 we will have 10 possible choices for 5th place . hence total number of combination 26*26*10*10*10=676000


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