Help me solving these (1 Viewer)

[efini4eva]

32 digit number signature consisting of only ones and zeros

it can be 1001001010101110111000110 something like that...so order is not important i think.

I think it's combination in probability

a) A computer 'hacker' wants to creak his friend's signature. He knows that it has an equal number of ones and zeros. Find the number of possibilities that he must consider? ( 2 marks )



b) Find the probability that in any signature chosen at random. the last 5 digits are either all ones or all zeros. ( 2 marks )


And different question.

The letters of the word TRANSITION are arranged in a row. Find the probability that the letter A occurs somewhere to the right of the letter R. ( 4 mark )

 

[noah]

a) 32!/((16!)(16!))

b) 2*(32-5)!/((16!)((16-5)!))

= 2*(27)!/((16!)(11!))

 

[noah]

TRANSITION

9!+8*8!+7*8!+...+2*8!+8!

=8!(9+8+...+2+1)

=8!*45 possibilities

Because if R is the first letter there are 9 remaining letters to be arranged in any way.

if R is the second letter, A can be in 8 of the remaining spots and there are 8 lrtters left to arrange

if R is the third letter, A can be in 7 of the remaining spots and there are 8 letters left to arrange.

and so on


P=no. of posibilities over total arrangements

total arrangements= 10!/((2!)(2!))

P= (45*8!)(2!)(2!)/10!

though i could be wrong

though i could be wrong

 

[efini4eva]

Thanks!! noah!

 

[efini4eva]

i think total arrangements =10! / 2! x 2! x 2! as there are 2xT and 2xN 2xI

 

[noah]

oh yeah 2 I's didm't see that