Puzzles

Posted On 2016-11-22

Answer For 4 digit numbers above 3600 we have to take 2 cases because if 1st digit is 3, then 2nd digit can only be 6,7,8,9. However, if 1st digit is 4,5,6,7,8 or 9, then 2nd digit can be any number. Since the 2nd digit is dependent on what the 1st digit is we need to solve this problem using 2 cases Case 1: If 1st digit is 3 1st digit can be selected in 1 way. 2nd digit can be any digit out of 6,7,8,9 ie we can select 2nd digit in 4 ways. 3rd digit can be any digit from 1 to 9. However, we cannot select the 1st or 2nd digit. Hence, we can select the 3rd digit in 7 ways. Similarly 4th digit can be selected in 6 ways. Total ways= 1×4×7×6= 168 Case 2: If the first digit is 4,5,6,7,8 or 9 1st digit can be selected in 6 ways. Now, there is no restriction on 2nd digit. Hence, 2nd 3rd and 4th digit can be selected in 8, 7 & 6 ways respectively. Total ways= 6×8×7×6= 2016 ways Hence, total numbers possible= 168+2016= 2184 Note: Case 1 and 2 cannot happen simult

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