How many numbers of 7 digits can be formed from the digits 1,2,3,4,5,6,7 if there are not more than 2 digits between 1 and 2.
Given Ans - 15
Here 's my attempt -
Case - I
When 1 and 2 are consecutive - 6!*2!
Case - II
When 1 and 2 are alternate - 5!*5c1 * 2!
Case - III
when there are 2 digits between 1 and 2!
Total ways = CaseI + CaseII + Case III = 2400
Pls. let me know if my answer / approach is correct.
also, if anybody knows about a better approach pls. share with all.
I propose this:
Case I: fine 6! * 2!
Case II: consider a group of 3 number with 1 and 2 at the ends. Given that the middle number is a specific number, there are 5!x2 ways to arrange the numbers x 5 possible different numbers. Hence, IMO, your calculation for case II is fine also.
Case III. consider a group of 4 numbers with 1 and 2 at the ends. Given that the middle numbers are two specific numbers, there are 4!x2 ways to arrange the numbers. However, there are 5C2 ways to choose the two middle numbers AND 2 ways to arrange them
. Hence, your calculation for case III in my opinion is off by a factor of 2.
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993