# Answers and Solutions

Question: Is there any other method to solve Question 2 other than (all Cases-not wanted cases)

Posted by: Priyanshi S. on 01.06.2021

Case 1 : Exactly one digit repeated twice --

eg AABCD     Words= 10C1 X 9C3 X {5!/2!} =50400

Case 2 : Exactly one digit repeated thrice --

eg AAABC     Words= 10C1 X 9C2 X {5!/3!}= 7200

Case 3 : Exactly one digit repeated four times --

eg AAAAB     Words= 10C1 X 9C1 X {5!/4!}= 450

Case 4 : Exactly one digit repeated five times --

eg AAAAA     Words= 10C1 X {5!/5!}= 10

Case 5 : Exactly two digits repeated two times each --

eg AABBC     Words= 10C2 X 8C1 X {5!/(2! 2!)}= 10800

Case 6 : Exactly two digits repeated one of them twice and the other thrice --

eg AAABB     Words= 10C1 X 9C1 X {5!/(2! 3!)}= 900

FINAL ANSWER= Sum of all cases= 69760

Number of words with no digit repeated= 10x9x8x7x6= 30240

Total five letter words formed using 10 letters =10x10x10x10x10= 10^5

Number of words which have at least one letter repeated = Total words - Number of words with no digit repeated =

10^5 - 30240 = 69760

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