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an n digit no is apositive no. with exactly n digit . Nine hundred distinct n digit no. r to be formed using only 3 digit 5,7.9. The smallest value of n is there r 4 balls of different clolur and 4 boxes colour same as those of balls the no of ways in which the ballon each 1 box could be plkaced such that any ball dosenot go to the box of its own colour


  1.  an n digit no is apositive no. with exactly n digit . Nine hundred distinct n digit no. r to be formed using only 3 digit 5,7.9. The smallest value of n is

  2. there r 4 balls of different clolur and 4 boxes colour same as those of balls the no of ways in which the ballon each  1 box could be plkaced such that any ball dosenot go to the box of its own colour


 

Grade:12

1 Answers

Sukhendra Reddy Rompally B.Tech Mining Machinery Engg, ISM Dhanbad
93 Points
14 years ago

Hi Anita Sharma!!!

1-the answer is 7,

explanation-u have the digits 5,7,9.and u have to form 900 distinct numbers.For any n digit number,the possible ways of filling each digit{that is units,tens,hundreds,etc} is 3

for a 1 digit number,total no. of ways is 3 power1=3{5,7,9}

for a 2 digit number,total no. of ways is 3 power 2+9{55,57,59,75,77,79,95,97,99}

for an n digit number,total no. of ways is 3 power n

now,we hav total no. of ways is 900,so we have to find a minimum n for which 3 power n is greater dan or equal to 900

3 power 6 is 729 and 3 power 7 is 2187,hence 7 is the answer

now the second ques

this is called dearrangements,and der is a forumla given in Dasgupta{which i dont remember now :(}

but,there is an alternative way,where u could choose cases where all 4 go to the ryt boxes,any 3,any2 and any 1 go into the right boxes and subtract the sum of all these cases from 4! which is the total no. of ways of arranging the balls{NOTE-THIS IS POSSIBLE ONLY ID THE NO. IS SMALL,LIKE 4 HERE,U CANT WASTE TYM TRYING TO SORT OUT CASES IF THE NO. OF BALLS/BOCES IS HIGH LIKE 7 OR 8,SO TRY TO CHECK THE FORMULA FOR DEARRANGEMENTS}

No. of ways off all balls goin into the ryt boxes is 1

No. of ways of any 3 balls going into the right boxes is zero as if 3 go into the right boxes,then the 4th will have to go into the right one

No. of ways of any 2 balls goin into the ryt box is 4c2 * 1! = 6*1=6{i.e; choosin 2 go into the ryt ones,and the total ways others go into the wrong ones,which is (2-1)!}

No. of ways of any 1 ball going into the ryt box is 4c1 * 2! =8

now,the sum of all these is 1+0+6+8=15

total ways of arranging is 4!=24

so,total ways of all balls going into the wrong boxes is 24-15 =9

Hope u understand the solutions

If u dont,feel free to call me on 07209736303

ALL THE BEST

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