(Puzzles) Challenging Mathematical Puzzles (34)
Puzzles : Challenging Mathematical Puzzles
115. Somebody marked the six faces of a die with the numbers 1, 2 and 3 - each number twice. The die was put on a table. Four people - Abu, Babu, Calu and Dabu - sat around the table so that each one was able to see only three sides of the die at a glance.
* Abu sees the number 1 and two even numbers.
* Babu and Calu can see three different numbers each.
* Dabu sees number 2 twice and he can't remember the third number.
What number is face down on the table?
Answer : - Number 3 is face down on the table.
If Abu can see two even numbers i.e. number 2 twice, and if Dabu can see number 2 twice, then number 2 must be facing up.
Now everything else is simple. (see the following diagram)
Dabu
Abu
1
3 2 2
1
Calu
Babu
Thus, the number hidden from the view is number 3 and hence the answer.
116. Two identical pack of cards A and B are shuffled throughly. One card is picked from A and shuffled with B. The top card from pack A is turned up. If this is the Queen of Hearts, what are the chances that the top card in B will be the King of Hearts?
Answer : - 52 / 2703
There are two cases to be considered.
CASE 1 : King of Hearts is drawn from Pack A and shuffled with Pack B
Probability of drawing King of Hearts from Pack A = 1/51 (as Queen of Hearts is not to be drawn)
Probability of having King of Hearts on the top of the Pack B = 2/53
So total probability of case 1 = (1/51) * (2/53) = 2 / (51 * 53)
CASE 2 : King of Hearts is not drawn from Pack A
Probability of not drawing King of Hearts from Pack A = 50/51 (as Queen of Hearts is not to be drawn)
Probability of having King of Hearts on the top of the Pack B = 1/53
So total probability of case 2 = (50/51) * (1/53) = 50 / (51 * 53)
Now adding both the probability, the required probability is
= 2 / (51 * 53) + 50 / (51 * 53)
= 52 / (51 * 53)
= 52 / 2703
= 0.0192378
117. How many possible combinations are there in a 3x3x3 rubics cube?
In other words, if you wanted to solve the rubics cube by trying different combinations, how many might it take you (worst case senerio)?
How many for a 4x4x4 cube?
Answer : - There are 4.3252 * 10^19 possible combinations for 3x3x3 Rubics and 7.4012 * 10^45 possible combinations for 4x4x4 Rubics.
Let's consider 3x3x3 Rubics first.
There are 8 corner cubes, which can be arranged in 8! ways.
Each of these 8 cubes can be turned in 3 different directions, so there are 3^8 orientations altogether. But if you get all but one of the corner cube into chosen positions and orientations, only one of 3 orientations of the final corner cube is possible. Thus, total ways corner cubes can be placed = (8!) * (3^8)/8 = (8!) * (3^7)
Similarly, 12 edge cubes can be arranged in 12! ways.
Each of these 12 cubes can be turned in 2 different directions, so there are 2^12 orientations altogether. But if you get all but one of the edge cube into chosen positions and orientations, only one of 2 orientations of the final edge cube is possible. Thus, total ways edge cubes can be placed = (12!) * (2^12)/2 = (12!) * (2^11)
Here, we have essentially pulled the cubes apart and stuck cubes back in place wherever we please. In reality, we can only move cubes around by turning the faces of the cubes. It turns out that you can't turn the faces in such a way as to switch the positions of two cubes while returning all the others to their original positions. Thus if you get all but two cubes in place, there is only one attainable choice for them (not 2!). Hence, we must divide by 2.
Total different possible combinations are
= [(8!) * (3^7)] * [(12!) * (2^11)] / 2
= (8!) * (3^7) * (12!) * (2^10)
= 4.3252 * 10^19
Similarly, for 4x4x4 Rubics total different possible combinations are
= [(8!) * (3^7)] * [(24!)] * [(24!) / (4!^6)] / 24
= 7.4011968 * 10^45
Note that there are 24 edge cubes, which you can not turn in 2 orientations (hence no 2^24 / 2). Also, there are 4 center cubes per face i.e. (24!) / (4!^6). You can switch 2 cubes without affecting the rest of the combination as 4*4*4 has even dimensions (hence no division by 2). But pattern on one side is rotated in 4 directions over 6 faces, hence divide by 24.
118. There are 20 people in your applicant pool,
including 5 pairs of identical twins.
If you hire 5 people randomly, what are the chances you will hire at least 1
pair of identical twins?
Answer
The probability to hire 5 people with at least 1 pair of identical twins is
25.28%
5 people from the 20 people can be hired in 20C5 = 15504 ways.
Now, divide 20 people into two groups of 10 people each :
G1 - with all twins
G2 - with all people other than twins
Let's find out all possible ways to hire 5 people without a single pair of
indentical twins.
People from G1 |
People from G2 |
No of ways to hire G1 without a single pair of indentical twins |
No of ways to hire G2 |
Total ways |
0 |
5 |
10C0 |
10C5 |
252 |
1 |
4 |
10C1 |
10C4 |
2100 |
2 |
3 |
10C2 * 8/9 |
10C3 |
4800 |
3 |
2 |
10C3 * 8/9 * 6/8 |
10C2 |
3600 |
4 |
1 |
10C4 * 8/9 * 6/8 * 4/7 |
10C1 |
800 |
5 |
0 |
10C5 * 8/9 * 6/8 * 4/7 * 2/6 |
10C0 |
32 |
Total |
11584 |
Thus, total possible ways to hire 5 people without a single pair of identical
twins = 11584 ways
So, total possible ways to hire 5 people with at least a single pair of
identical twins = 15504 - 11584 = 3920 ways
Hence, the probability to hire 5 people with at least a single pair of identical
twins
= 3920/15504
= 245/969
= 0.2528
= 25.28%