Q.1 How many numbers are there between 100 and 1000 such that every digit is either 2 or 9?

Q.2 In how many ways can the following prizes be given away to a class of 30 students, 1st and 2nd in Maths, 1st and 2nd in physics, 1st in chem and 1st in eng

please explain me these questions and also its answers......

1.

the numbers between 100 and 1000 are clearly the three digit number.

all the three digits can be filled by 2 ways , i.e. either 2 or 9.

thus the total number of ways = 2*2*2 = 8  ways

2.

 

total number of students = 30

total number of first prizes are 4 (maths, physics , chemistry and english)

total number of second prizes are 2 (maths, physics)

first prize can be given to any of the 30 students.

since the student , who got the first prize in maths , may also have the first prize in physics , chemistry and / or english.

4 first prizes can be given as 30*30*30*30 ways = 810000

2nd prize can be given to 29 students (excluding the student who got 1st prize)

the student , who got the first prize in maths , may also have the 2nd prize in physics, or chemistry or english.

similarly the student who got the first prize in physics, may also have the 2nd prize in any of the subject other than physics.

2 2nd prizes can be given as 29*29 ways = 841

therefore total number of required ways = 30*30*30*30*29*29 = 681210000

if the condition is given that : a student may have only one prize.

then total number of prizes are: 4+2 = 6

total number of ways = 30*29*28*27*26*25 

 

hope this helps you.

cheers!!

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