permutation

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

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!!!!!!

one_pacifist wrote: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

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

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

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

721tjm wrote: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