How many strings of 8 letters from the English alphabet are there where... a. th

Question

How many strings of 8 letters from the English alphabet are there where...

a. there is EXACTLY one "a" and letters CANNOT be repeated?

b. there is an x and y and the x must come before the y (to the left of y) and no letters can be repeated?

c. there is EXACTLY one "a" and letters can be repeated?

d. there is the occurrence of "xyz" consecutively and letters cannot be repeated?

e. there are exactly 3 x's and exactly 4 y's?

f. the string is the same if it is written in reverse order (a palindrome)?

Explanation / Answer

How many strings of 8 letters from the English alphabet are there where...

a. there is EXACTLY one "a" and letters CANNOT be repeated? THERE ARE 26 ALPHABETS A IS COMPULSORY ..SO WE HAVE TO SELECT 7 MORE LETTERS FROM 25 THIS CAN BE DONE IN C[25,7] WAYS...AS THERE IS NO REPETETION..... THE 8 LETTERS CAN THEN BE PERMUTATED IN 8! WAYS .. SO TOTAL NUMBER OF WORDS ARE = [C(25,7)][8!]=[{(25!)}{(8!)}] / [(7!)*(18!)] =[(25!)*8] / [(18!)] b. there is an x and y and the x must come before the y (to the left of y) and no letters can be repeated?
6 LETTERS CAN BE SELECTED FROM 26-2=24 IN C[24,6] WAYS X HAS TO COME BEFORE Y .. SO X CAN BE IN PLACE.1..THEN Y CAN BE IN ANY OF THE 7 REMAINING PLACES ...7 WAYS X CAN BE PLACE 2 .....THEN Y CAN BE IN ANY OF THE 6 REMAINING PLACES......6 WAYS .. ..ETC.... X CAN BE IN PLACE 6....THEN Y CAN BE IN 1 REMAINING PLACE .......1 WAYS SO X AND Y CAN BE PLACED IN 7! WAYS .... REST OF 6 LETTERS CAN BE PERMUTATED IN 6! WAYS SO TOTAL NUMBER OF WAYS ARE = [C(24,6)][7!][6!] = [(24!)(7!)]/[(18!)] c. there is EXACTLY one "a" and letters can be repeated? SO LET US PUT A IN ANY ONE OF THE 8 POSITIONS IN 8 WAYS.. REST 7 POSITIONS CAN BE FILLED BY ANY ONE OF THE 25 IN 25 WAYS EACH .. SO TOTAL NUMBER OF WAYS =8([25*25*...7 TIMES]= 8*[25^7] d. there is the occurrence of "xyz" consecutively and letters cannot be repeated? XYZ AS A COMBINATION CAN START FROM PLACE 1 TO PLACE 6 .....6 WAYS ..... REST 5 LETTERS CAN BE SELECTED FROM 23 IN C[23,5] WAYS .... THESE 5 CAN BE PERMUTATED IN 5! WAYS. SO TOTAL NUMBER OF WAYS = [6]*[C(23,5)]*[5!] = [6][23!] / [18!] e. there are exactly 3 x's and exactly 4 y's? THAT LEAVES US ONLY ONE PLACE WHICH CAN BE FROM 24 LETTERS IN ...24 WAYS... IF YOU WANT DIFFERENT STRINGS THEN 3 X'S , 4 Y'S , AND ONE ANOTHER LETTER CAN BE PERMUTATED IN [8!]/[(4!)(3!)] SO WE HAVE IN ALL [24][8!] / [(4!)(3!)] f. the string is the same if it is written in reverse order (a palindrome)? NO MENTION OF REPETETIONS.. ASSUMING ONLY ONE REPETETION OF EACH LETTER .....THAT IS WE CAN HAVE ONLY 2A'S OR 2B'S ..ETC... SO ..PLACES ....1&8 ; 2&7 ; 3&6 ; 4&5 ; SHALL BE OCCUPIED BY IDENTICAL LETTERS . THAT IS WE CAN SELECT 4 LETTERS FROM 26 IN C[26,4] WAYS THEY CAN THEN BE PERMUTATED IN 4! WAYS.. SO TOTAL NUMBER OF POSSIBILITIES ARE ..= [(4!)(26!)] / [(4!)(22!)] How many strings of 8 letters from the English alphabet are there where...

a. there is EXACTLY one "a" and letters CANNOT be repeated? THERE ARE 26 ALPHABETS A IS COMPULSORY ..SO WE HAVE TO SELECT 7 MORE LETTERS FROM 25 THIS CAN BE DONE IN C[25,7] WAYS...AS THERE IS NO REPETETION..... THE 8 LETTERS CAN THEN BE PERMUTATED IN 8! WAYS .. SO TOTAL NUMBER OF WORDS ARE = [C(25,7)][8!]=[{(25!)}{(8!)}] / [(7!)*(18!)] =[(25!)*8] / [(18!)] b. there is an x and y and the x must come before the y (to the left of y) and no letters can be repeated?
6 LETTERS CAN BE SELECTED FROM 26-2=24 IN C[24,6] WAYS X HAS TO COME BEFORE Y .. SO X CAN BE IN PLACE.1..THEN Y CAN BE IN ANY OF THE 7 REMAINING PLACES ...7 WAYS X CAN BE PLACE 2 .....THEN Y CAN BE IN ANY OF THE 6 REMAINING PLACES......6 WAYS .. ..ETC.... X CAN BE IN PLACE 6....THEN Y CAN BE IN 1 REMAINING PLACE .......1 WAYS SO X AND Y CAN BE PLACED IN 7! WAYS .... REST OF 6 LETTERS CAN BE PERMUTATED IN 6! WAYS SO TOTAL NUMBER OF WAYS ARE = [C(24,6)][7!][6!] = [(24!)(7!)]/[(18!)] c. there is EXACTLY one "a" and letters can be repeated? SO LET US PUT A IN ANY ONE OF THE 8 POSITIONS IN 8 WAYS.. REST 7 POSITIONS CAN BE FILLED BY ANY ONE OF THE 25 IN 25 WAYS EACH .. SO TOTAL NUMBER OF WAYS =8([25*25*...7 TIMES]= 8*[25^7] d. there is the occurrence of "xyz" consecutively and letters cannot be repeated? XYZ AS A COMBINATION CAN START FROM PLACE 1 TO PLACE 6 .....6 WAYS ..... REST 5 LETTERS CAN BE SELECTED FROM 23 IN C[23,5] WAYS .... THESE 5 CAN BE PERMUTATED IN 5! WAYS. SO TOTAL NUMBER OF WAYS = [6]*[C(23,5)]*[5!] = [6][23!] / [18!] e. there are exactly 3 x's and exactly 4 y's? THAT LEAVES US ONLY ONE PLACE WHICH CAN BE FROM 24 LETTERS IN ...24 WAYS... IF YOU WANT DIFFERENT STRINGS THEN 3 X'S , 4 Y'S , AND ONE ANOTHER LETTER CAN BE PERMUTATED IN [8!]/[(4!)(3!)] SO WE HAVE IN ALL [24][8!] / [(4!)(3!)] f. the string is the same if it is written in reverse order (a palindrome)? NO MENTION OF REPETETIONS.. ASSUMING ONLY ONE REPETETION OF EACH LETTER .....THAT IS WE CAN HAVE ONLY 2A'S OR 2B'S ..ETC... SO ..PLACES ....1&8 ; 2&7 ; 3&6 ; 4&5 ; SHALL BE OCCUPIED BY IDENTICAL LETTERS . THAT IS WE CAN SELECT 4 LETTERS FROM 26 IN C[26,4] WAYS THEY CAN THEN BE PERMUTATED IN 4! WAYS.. SO TOTAL NUMBER OF POSSIBILITIES ARE ..= [(4!)(26!)] / [(4!)(22!)]

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