Permutations 4

Question

Problem: questions on exam
Level: Easy/Medium

There are six questions in a question paper. In how many ways can a student solve one or more questions?

tkarthi4u · Answer

The total number of arrangements = 6!

but you are including one ways that he will never solve any questions.

so the Ans . (6!-1)

LenaA · Answer

why is it 6!-1?
He has 6 questions to solve. There are two possible outcomes for each question Correct or Mistake. 
When he answes 2 questions, there are 2X2 = 4 ways: CM, MC, CC, MM
When he ansers 6 questions there are 2^6=64
But one of the oucomes is MMMMMM , so in total 63 ways 
Am I wrong?

tkarthi4u · Answer

i think you are perfectly right. i made a big blunder.

6! = will definitely give me the number of ways i can arrange 6 six questions. not solving them

thanks.

flyingbunny · Answer

This one is correct
2^6-1

rohansherry · Answer

But it dint say in ow many ways you can get the answer...it says how many ways you can solve it...a lil confusion here ....What is the OA

LenaA · Answer

6! refers to the way you can arrange six questions. However, the problem does not ask in how many ways you can arrange them on the exam paper. It asks about the SOLUTION to a given exam ...CMMMMM or CCMMMM, etc.

bhushan252 · Answer

look at it this way

each question has 2 choices....answer or leave it...so 6 questions will have 2^6 choices with one choice of leaving all questions so minus 1 ....hence 2^6 -1

rohansherry · Answer

guess, makes more sense......

CasperMonday · Answer

Guys,

the official answer is 2^6-1

since each question can either be "solved" or "not solved", there are two ways to deal with each question
since we have 6 questions, the total number of ways turns to be 2*2*2*2*2*2=2^6
now the problem says that a student has to solve at least one question. this means that we should take out a variant when he fails to answer all questions, that's why 2^6-1

rohansherry · Answer

There are 6 cases
1) when he answers only 1 correct - total ways = 6C1
2) when he answers only 2 correct - total ways = 6C2
3) when he answers only 3 correct - total ways = 6C3
4) when he answers only 4 correct - total ways = 6C4
5) when he answers only 5 correct - total ways = 6C5
6) when he answers only 6 correct - total ways = 6C6

Hence total ways = 6C1 + 6C2 + 6C3 + 6C4 + 6C5 + 6C6 = 32 ways

I misunderstood..???/

LenaA · Answer

He has 6 questions to solve. There are two possible outcomes for each question Correct or Mistake. 
When he answes 2 questions, there are 2X2 = 4 ways: CM, MC, CC, MM
When he ansers 6 questions there are 2^6=64
But one of the oucomes is MMMMMM , so in total 63 ways 
Am I wrong?

But it dint say in ow many ways you can get the answer...it says how many ways you can solve it...a lil confusion here ....What is the OA

6! refers to the way you can arrange six questions. However, the problem does not ask in how many ways you can arrange them on the exam paper. It asks about the SOLUTION to a given exam ...CMMMMM or CCMMMM, etc.

There are 6 cases
1) when he answers only 1 correct - total ways = 6C1
2) when he answers only 2 correct - total ways = 6C2
3) when he answers only 3 correct - total ways = 6C3
4) when he answers only 4 correct - total ways = 6C4
5) when he answers only 5 correct - total ways = 6C5
6) when he answers only 6 correct - total ways = 6C6

Hence total ways = 6C1 + 6C2 + 6C3 + 6C4 + 6C5 + 6C6 = 32 ways

I misunderstood..???/

I do not understand how do you get 32 ways?

6C2 = 6C4 =\frac{6!}{4!	imes2!}=15 
 6C1=6C5=\frac{6!}{5!	imes1!}=6 
 6C3=\frac{6!}{3!	imes3!}=20 
6C6=1

Total=15	imes2+6	imes2+20+1=63 
Your way is correct, something with calculation wrong...

Permutations 4

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