Trillian's mice III
Trillian keeps pet mice. At least two are white, and at least two are black.
If four mice are chosen at random (without replacement), the probability that two are white is equal to the probability that none are white.
Given further that Trillian has more mice than toes (of which she has the usual ten), what's the least number of mice she could possibly have?
NB. This is the final—and hardest—problem in a series of three. You are encouraged to solve the previous problems first.
The answer is 15.
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3 solutions
You could also write this as a quadratic in w and then find the discriminant.
I used the fact that b-2 is either divisible by w or w-1
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but what about 6? don't forget about the 6. it could be that b-2 is divisible by 6 but not divisible by w or w-1.
Alternatively, note that w must be even, and that 3(w-1)-1=2w, which implies that w=4.
Did the same way
It would have been a little tougher if Trillian had more than 1 5 toes. I get 3 5 white mice and 8 7 black mice as the next solution.
(You can convert your equation to a Pell equation ( 2 b − 5 ) 2 − 6 ( 2 w − 1 ) 2 = − 5 .)
Since Trillian has only white and black mice, if she picks up four mice and has not picked up a white mouse, she has to have picked up four black mice in a row. If b represents the number of black mice and t represents the total number of mice, this means that the probability she picked up no white mice is: (b)(b-1)(b-2)(b-3) / t(t-1)(t-2)(t-3)
If Trillian picks up two white mice, that means that she also picked up two black mice. There are six ways that she could have picked up two white mice. If her first and second mice were white, her second and third mice were white, her third and fourth mixe were white, her first and third mice were white, her second and fourth mice were white and her first and fourth mice were white. If w represents the number of white mice, the probability she picked up two white mice is: 6(b)(b-1)(w)(w-1) t(t-1)(t-2)(t-3) Since we are told the probabilities are equal to each other, we get:
6(b)(b-1)(w)(w-1) / t(t-1)(t-2)(t-3)= (b)(b-1)(b-2)(b-3)/t(t-1)(t-2)(t-3) If we multiply both sides by t(t-1)(t-2)(t-3) and divide both sides by b(b-1), we get:
6w(w-1)=(b-2)(b-3) Since w, w-1, b-2 and b-3 are integers, we know that one of the following possibilities is true. The factors of six are somehow split between w and w-1 to get b-2 and b-3. a. 6w= b-2 w-1= b-3
b. 6w= b-3 w-1= b-2
c. w= b-2 6(w-1)= b-3
d. w= b-3 6(w-1)= b-2
e. 2w= b-2 3(w-1)= b-3
f. 2w= b-3 3(w-1)= b-2
g. 3w=b-2 2(w-1) = b-3
h. 3w=b-3 2(w-1) = b-2
The first four possibilities can immediately be ruled out, since six times a number and one times a consecutive number cannot equal to consecutive natural numbers.
We can see through some work that the only way we can w+b>10, still fulfill one of the last four equations and still get the smallest answer possible is if we set w= 4 and b = 11. We end up fulfilling f. Therefore, the number of rabbits is 15.
Call n is number of white mice, s is total number of mice. We have: 2 ! × ( n − 2 ) ! n ! × 2 ! × ( s − n − 2 ) ! ( s − n ) ! = 4 ! × ( s − n − 4 ) ! ( s − n ) ! So: s 2 − s × ( 2 × n + 5 ) − 5 × n 2 + 1 1 × n + 6 = 0 ( 1 ) The smallest n that satisfies (1) and more than 2 is 4. s=15
Denote by w the number of white mice and by b the number of black ones. We require that: ( 4 w + b ) ( 2 w ) ( 2 b ) = ( 4 w + b ) ( 4 b )
Simplifying this gives the constraint:
6 w ( w − 1 ) = ( b − 2 ) ( b − 3 )
A quick brute force search yields w = 4 and b = 1 1 as a viable solution.