The Great Score Of 100

Adarsh, the great batsman, is batting. His end score is 100 not out. Being the best batsman on the team, he scores only in 4s and 6s. In how many ways can he achieve this score?


The answer is 9.

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3 solutions

Rishabh Jain
Mar 6, 2016

This is just the starting ... India will also emerge as the World Cup T20 Champions 2k16....

Anyways here's a solution... Let the no. Of 4 s 4's and 6 s 6's hit by Adarsh be x x and y y respectively. Then:- 4 x + 6 y = 100 2 x + 3 y = 50 x = 25 3 y 2 4x+6y=100\\ \implies 2x+3y=50 \\ \implies x=25-\dfrac{3y}{2} From here we can see that since x is a positive integer y must be a even integer such that y [ 0 , 16 ] y\in [0,16] i.e 9 \boxed 9 values. Correspondingly we'll get 9 \boxed 9 values of x x too i.e 9 \boxed 9 values.

Yes, good solution :)

And sure, India will ;)

Mehul Arora - 5 years, 3 months ago

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Yeah ... Right

Rishabh Jain - 5 years, 3 months ago

Nicely done!

Adarsh Kumar - 5 years, 3 months ago

Great to see so many cricket fans... All hail India!

Rishik Jain - 5 years, 3 months ago
Rajen Kapur
Mar 17, 2016

In my view the case of zero 6's should be excluded, as then only 4's and not 4's and 6's as per the problem statement.

But sir, nowhere is it stated that he cannot score only in 4s. It is only mentioned that he scores in 4s and 6s.

Mehul Arora - 5 years, 3 months ago
Kyle Misiaszek
Mar 16, 2016

I came up with 525456 as an answer. The suggested solutions so far are assuming that there is only one way to get 22 4's and 2 6's. If you consider the number of unique arrangements of 22 4's and 2 6's, there are actually 276 ways to get 22 4's and 2 6's to sum to 100. I feel that my solution is correct because scoring 4 the first 22 times at bat then scoring 6 the remaining two times at bat would be recorded differently (chronologically) in a score book than would scoring 6 the first two at bats and scoring 4 for the remaining 22 at bats.

# of 4’s # of 6’s # of ways for this row
25 0 25C25 = 1
22 2 24C22 = 276
19 4 23C19 = 8855
16 6 22C16 = 74613
13 8 21C13 = 203490
10 10 20C10 = 184756
7 12 19C7 = 50388
4 14 18C4 = 3060
1 16 17C1 = 17

Total: 525456

525456 is the incorrect answer, How did you get it right?

Also, this solution is incorrect. refer to Rishabh's solution above.

Mehul Arora - 5 years, 3 months ago

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Mehul, one of the wonderful things about mathematics is that there are sometimes multiple paths to an answer and also different perspectives to a problem.

I never said Rishabh was wrong. I never said I was right. Me feeling right isn't making a claim that my work is absolutely and definitively correct.

So I challenge you to look at this problem and my previous post with an open mind. Leave your previous assumptions behind and consider if my perspective/assumption is also a valid viewpoint of this problem.

Should unique orderings of the addends be counted or not?

Kyle Misiaszek - 5 years, 3 months ago

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I'm sorry sir. I realise I have been a jerk before :P

I think unique ordering should not be counted in this case.

Mehul Arora - 5 years, 3 months ago

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@Mehul Arora Mehul,

You don't need to apologize. I wasn't offended by anything you wrote.

The world is full of people with different opinions on math, politics, religion, etc. and that's what makes the world an exciting and interest place. I'm just encouraging anybody who reads this page to keep an open mind.

I'm also going to pose a new problem to think about:

Every year I play in a recreational softball league. The league has 10 teams and every team plays every other team exactly once in the regular season. For the last 7 years that I've played, my team has finished with 6 wins and 3 losses each and every year. My newest season is about to start and I wouldn't be surprised if my team again finished 6-3 (6 wins, 3 losses). The schedule has been made and I know which week I'm playing which team. Let's say that I'm playing team A in week 1 and team B in week 2 and so on. In how many different ways could my team finish 6-3?

Kyle Misiaszek - 5 years, 3 months ago

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