Solved Examples

[ravihanda]

Seven different objects may be divided among three people.In how many ways can this be done if at least one of them gets exactly one object?
a)2484 b)1176 c)729 d)2424 e)none of these

Case 1: When only one person has exactly one object
3C1*7C1*(6C2+6C3+6C4) = 1050
Case 2: When two person have exactly one object
7C2*2!*3 = 126
Total ways = 1050+126 = 1176. Option B

[ravihanda]

There are 7 pairs of black shoes and 5 pairs of white shoes. They are all put into a box and shoes are drawn one at a time. To ensure that at least one pair of black shoes are taken out, what is the number of shoes required to be drawn out?
a) 12 b) 13 c) 7 d)18

We need to consider the worst case scenario in such cases. First take out all the white shoes i.e. 10 shoes. Then take out all the left foot black shoes i.e. 7 shoes. Till now, we have taken out 17 shoes and we don't have a black pair. The 18th shoe we take out, will give us on pair of black shoes. Option D

[ravihanda]

How many numbers smaller than 2.10^8 & r divisible by 3 can be written by means of the digits 0,1 and 2 (exclude single digit and double digit numbers) ?
Numbers which are less than 2x10^8 and can be formed by 0,1 and 2 = 2x3^8 = 13122 (First position can be filled with 0 or 1 and the other eight positions can be filled with 0 or 1 or 2).
one digit numbers are 3 and two digit numbers are 3^2 = 9. Remove these 12 numbers and we are left with 13110. Out of these every third one would be divisible by 3. So the answer would be 13110/3 = 4370.
The answer in my edition of Arun's book is given as 294840. I guess that is a printing mistake.

[ravihanda]

Find the sum of all the four digit numbers that can be formed with the digits 3,2,3,4.
a. 40982 b. 39996 c. 41682 d.none of these

No. of four digit numbers that would end with 2 = 3
No. of four digit numbers that would end with 4 = 3
No. of four digit numbers that would end with 3 = 6
Sum of the digits at the unit's place for all the four digit numbers = 2x3 + 4x3 + 3x6 = 6 + 12 + 18 = 36
Same pattern would be followed for other positions as well.
Sum of the digits at the ten's place = 36
Sum of the digits at the hundred's place = 36
Sum of the digits at the thousand's place = 36
Total sum = 36x1000 + 36x100 + 36x10 + 36 = 39996. Option B

[bharadwaj12]

How many 4-digit numbers are there whose decimal notation contains not more than 2 distinct digits?

Single Digit Numbers = 9

Two Digit Numbers (Excluding 0) = 9C2 x (2^4 - 2) = 36 x 14 = 504
9C2 because I need to select the two digits (Say a and b)
2^4 because at each position, it could be a or b.
-2 because I do not want aaaa or bbbb

Two Digit Numbers (Including 0) = 9 x (2^3 - 1) = 63
9 to select the digit other than 0(say a); 2^3 because it could be a or 0 at each position and -1 to eliminate 0000.

Total numbers = 9 + 504 + 63 = 576.
So, the answer is 576. Option B.

Sir I am not sure if I can reply here if its against rules please delete it sir..

Sir can u explain the second case here.. the (2^3-1) part.. There are four position since first position cant be zero the first position can be filled in 1 way(one amoung the selected 9 digits). Second,third and fourth postion can be filled in 2^3 ways. I dont understand why to subtract 1 as we will not get 0000 as per my post..

I manually wrote the combination and got the correct answer xxx0,xx00,x000,x0xx,xx0x,xx0x,x00x where we can select x in 9 ways so 9x7..But know what was wrong in my earlier, Plz tell me where I am going wrong sir.

Thx in advance

[ravihanda]

Dear Bharadwaj,

Relax. There are no such hard and fast rules For future reference, please go ahead with your posts, etc. I will move / delete / edit them if I think they are not useful for other readers.

I manually wrote the combination and got the correct answer xxx0,xx00,x000,x0xx,xx0x,xx0x,x00x where we can select x in 9 ways so 9x7..But know what was wrong in my earlier, Plz tell me where I am going wrong sir.

I guess you missed out on xxxx. The question says not more than two digits, so a number like 7777 is valid.