Permutations and Combinations Test - 6

To form a four digit number  with distinct digits we can use any four digits from the digits 0,1,2,3,4,5,6,7,8,9 without repetition 

Since 0 cannot come as the first digit of a four digit number( then it will be a three digit number) the first place can be filled by any of the 9  digits other than 0.

Now we have 9 more digits  left and since  repetition is not allowed the second place can by filled by any of these 9 digits

Similarly the third and fourth can be filled by any of the  8 and 7 digits respectively

Hence we get the four  places can be filled  together by 9×9×8×7=4536 different ways.

Since a  student can solve each question in 3 different ways - either   he can attempt the  first alternative ,or the second alternative or he can leave it unanswered.

Hence number of ways in which a student can attempt one or more of the 12 given questions =3¹²

Now we can consider a case that the student leave all the 12 given questions unanswered.

The number of ways, in which a student can select one or more questions out of 12 each having an alternative= 3¹²−1

for first place we have 20 students, for second we have 19 and for the third we have 18

²⁰P₃=20*19*18

First we will find the  number of three digit numbers (i.e, numbers from 100 to 999)which can be formed using the digits 0,1,2,3,4,5,6,7,8and 9 with repitition allowed .

Now we have the first place can be filled by any of the 9 digits other than 0 and since repetition is allowed the second and third can be filled by any of the ten digits.

Hence the total number of three digit numbers will be = 9×10×10=900

Now we will consider the case that the number does not have the digit 5.

Now the  first place can be filled by any of the 8 digits other than 0 and 5 and since repetition is allowed the second and third can be filled by any of the 9 digits other than 5.

Hence the total number of ways we can form a  three digit number with out 5 will be =8×9×9=648

Therefore the number of three digit numbers with atleast one 5=900−648=252

If there are n objects to be arranged in circular order the no of permutations possible=(n−1)!

First we will make the 5 girls around the table and this can be done in (5−1)!=4!=24 , different ways

Now  we have 5 places available between these girls and the 5 boys can be seated in these 5 available places in 5!=120, different ways

Hence the 5 boys and 5 girls can be arranged in 4!.5!=24.120=2880ways

First we will find the number of four digit numbers that can be formed using the digits 0,1,2,3,4,5,6,7,8,9  with repetition .

The first place can be filled by any of the 9 digits other than 0, and the second, third and the fourth places each  can be filled by any of the ten digits

Hence the total number of ways of forming a four digit number = 9×10×10×10=9000

Now we will find the number of four digit numbers in which nall the digits are distinct 

The first place can be filled by any of the 9 digits other than 0, and the second, can be filled by any of the remaining 9 digits since repetition is not possible

Similarly  third and the fourth places each  can be filled by 8 and 7 digits respectively

Hence the total number of ways of forming a four digit number  with distinct digits b= 9×9×8×7=4536

The total number of numbers from 1000 to 9999 (both inclusive) that do not have 4 different digits=9000−4536=4464