permutation

[gmatapril]

What is the difference between permutations and combinations.

Thanks.

[maihuna]

Its all about whether order matter or nat. You may pick any elementry book to see concepts.

[one_pacifist]

hi all

new to permutations and combinations topic, so need help on this ....

q) A set of 6 questions contains true /false type questions.
Maximum how many students can take the test if all the students answer differently from others and must attempt all the questions ?

a gud explanation on answer will help a lot!!!!!!

[Night reader]

hi all

new to permutations and combinations topic, so need help on this ....

q) A set of 6 questions contains true /false type questions.
Maximum how many students can take the test if all the students answer differently from others and must attempt all the questions ?

a gud explanation on answer will help a lot!!!!!!

Students can answer all 'True' --> 1
Students can answer all 'False' --> 1
Students can answer three 'True' the rest 'False' --> 6C3=6!/(3!*3!)=20
Students can answer four 'True' the rest 'False' --> 6C4=6!/(4!*2!)=15
Students can answer four 'False' the rest 'True' --> 6C4=6!/(4!*2!)=15
Students can answer five 'True' the rest 'False' --> 6C5=6!/(5!*1!)=6
Students can answer five 'False' the rest 'True' --> 6C5=6!/(5!*1!)=6

1+1+20+15+15+6+6=64 students can take the test if all the students answer differently from others and must attempt all the questions

[one_pacifist]

@ night reader thanks man.. for your prompt reply... and the explanation.... keep in touch... :mrgreen: [/img]

[721tjm]

I think this one might just be 2^6 = 64...

How many possible answers are there for the first question? 2
...second question? 2
and so forth

Multiply all the different scenarios together to equal 2*2*2*2*2*2 = 2^6 = 64

I think Night reader's on the right track but maybe double-counted a couple possibilities (the bold statements are the same scenario, the underlined are the same scenario, and the italics are the same scenario)

Students can answer all 'True' --> 1
Students can answer all 'False' --> 1
Students can answer two 'True' the rest 'False' --> 6C2=6*5/2=15
Students can answer two 'False' the rest 'True' --> 6C1=6*5/2=15
Students can answer three 'True' the rest 'False' --> 6C3=6!/(3!*3!)=20
Students can answer three 'False' the rest 'True' --> 6C3=6!/(3!*3!)=20
Students can answer four 'True' the rest 'False' --> 6C4=6!/(4!*2!)=15
Students can answer four 'False' the rest 'True' --> 6C4=6!/(4!*2!)=15
Students can answer five 'True' the rest 'False' --> 6C5=6!/(5!*1!)=6
Students can answer five 'False' the rest 'True' --> 6C5=6!/(5!*1!)=6

[Night reader]

thanks, I've over-counted/just edited my solution
I see your logic 6 ways to arrange each has 2 choices 2*2*...=2^6

I think this one might just be 2^6 = 64...

How many possible answers are there for the first question? 2
...second question? 2
and so forth

Multiply all the different scenarios together to equal 2*2*2*2*2*2 = 2^6 = 64

I think Night reader's on the right track but maybe double-counted a couple possibilities (the bold statements are the same scenario, the underlined are the same scenario, and the italics are the same scenario)

Students can answer all 'True' --> 1
Students can answer all 'False' --> 1
Students can answer two 'True' the rest 'False' --> 6C2=6*5/2=15
Students can answer two 'False' the rest 'True' --> 6C1=6*5/2=15
Students can answer three 'True' the rest 'False' --> 6C3=6!/(3!*3!)=20
Students can answer three 'False' the rest 'True' --> 6C3=6!/(3!*3!)=20
Students can answer four 'True' the rest 'False' --> 6C4=6!/(4!*2!)=15
Students can answer four 'False' the rest 'True' --> 6C4=6!/(4!*2!)=15
Students can answer five 'True' the rest 'False' --> 6C5=6!/(5!*1!)=6
Students can answer five 'False' the rest 'True' --> 6C5=6!/(5!*1!)=6