Yup, 4 is correct.x(x + 1) = 4y(y + 1)4x² + 4x = 16y² + 16y(4y + 2)² - (2x + 1)²= 3(4y - 2x + 1)(4y + 2x + 3) = 3 = 1*3 = 3*1 = (-1)(-3) = (-3)(-1)So, we will get 4 solutionsYup, 343 is correct.Twelve female workers and ten male workers are to be assigned to work in one of four different departments of a company. In how many ways can this be done if each department gets at least one worker?
Twelve female workers and ten male workers are to be assigned to work in one of four different departments of a company. In how many ways can this be done if each department gets at least one worker?
Let a,b,c,d be the 4 departments. Now, a+b+c+d = 22
Considering the natural number solutions for the equation, we have: No. of ways = (12+10-1)C(4-1) = 21C3 = 1330
Twelve female workers and ten male workers are to be assigned to work in one of four different departments of a company. In how many ways can this be done if each department gets at least one worker?
There are 22 members in total, so they be assigned to the departments in 4^22 ways, but this includes cases like exactly one department is one empty, exactly two are empty and where exactly three are empty which sums upto 12
Let a,b,c,d be the 4 departments.Now, a+b+c+d = 22Considering the natural number solutions for the equation, we have:No. of ways = (12+10-1)C(4-1) = 21C3 = 1330
Why are you considering 12 females identical and 10 males identical ??
@chillfactor : Since each department needs at least one worker and nothing is mentioned about selecting either male or female worker, that's why I considered them to be identical.
Yup, 4 is correct.x(x + 1) = 4y(y + 1)4x² + 4x = 16y² + 16y(4y + 2)² - (2x + 1)²= 3(4y - 2x + 1)(4y + 2x + 3) = 3 = 1*3 = 3*1 = (-1)(-3) = (-3)(-1)So, we will get 4 solutionsYup, 343 is correct.Twelve female workers and ten male workers are to be assigned to work in one of four different departments of a company. In how many ways can this be done if each department gets at least one worker?
4^22 total cases possible
4c1 way where only 1 is filled
4c2*2^22 ways where only 2 filled
4c3*3^22 ways where only 3 filled
so total -> 4^22 - 4 - 4c2*2^22 - 4c3*3^22 ?? i could be all wrong here
@chillfactor : Since each department needs at least one worker and nothing is mentioned about selecting either male or female worker, that's why I considered them to be identical.
No, as in it matters which individual goes to which department, they are all distinct beings. ? :/
There are 22 members in total, so they be assigned to the departments in 4^22 ways, but this includes cases like exactly one department is one empty, exactly two are empty and where exactly three are empty which sums upto 12So 4^22 - 12 ?PS: Trying to overcome my fears of PnC
Each of the 12 cases would also have different distribution in themselves .
So think about accounting that. Approach is in the right direction.
@Brooklyn just a little bit of +/- mix-up with that soln there :).
4^22 total cases possible4c1 way where only 1 is filled4c2*2^22 ways where only 2 filled-->not only two, at least two.4c3*3^22 ways where only 3 filled--->at least three.so total -> 4^22 - 4 - 4c2*2^22 - 4c3*3^22 ?? i could be all wrong here
Twelve female workers and ten male workers are to be assigned to work in one of four different departments of a company. In how many ways can this be done if each department gets at least one worker?
Yup, 4 is correct.x(x + 1) = 4y(y + 1)4x² + 4x = 16y² + 16y(4y + 2)² - (2x + 1)²= 3(4y - 2x + 1)(4y + 2x + 3) = 3 = 1*3 = 3*1 = (-1)(-3) = (-3)(-1)So, we will get 4 solutionsYup, 343 is correct.Twelve female workers and ten male workers are to be assigned to work in one of four different departments of a company. In how many ways can this be done if each department gets at least one worker?
Any postive integer can be expressed as 3k,3k+1,3k+23k(3k+2)=4*3k*(3k+2)No solution(3k+1)*(3k+2)=4*(3k+1)*(3k+2)No solution3k(3k+1)=4*(3k+1)*(3k+2)K is not integer3k(3k+1)=4*(3k+2)*(3k+3)K is not integer(3k+1)*(3k+2)=4*(3k+2)*(3k+3)no solutionno solution with any case
X and y are different, you've taken them to be the same.
im nt gud at exclusion and inclusion to i dont know hw to use it , i thnk dats y im missing some cases in mine
You're not missing cases. See when you do 4^22, it includes cases where one, two, three depts might be empty. Similarly, when you do 4C3*3^22 it includes cases where one or two depts might be empty.
You're not missing cases. See when you do 4^22, it includes cases where one, two, three depts might be empty. Similarly, when you do 4C3*3^22 it includes cases where one or two depts might be empty.
im nt gud at exclusion and inclusion to i dont know hw to use it , i thnk dats y im missing some cases in mine
Think set theory. For 2 sets, total is (1 at a time ) 0 (2 at a time). For 3 sets, (1 at a time) - (2 at a time) + (3 at a time). For 4 sets, (1 at a time) - (2 at a time) + (3 at a time) - (4 at a time) and so on...alternate +/- is the key to remember.
Also in this case the coefficients will be 4C4, 4C3, 4C2 etc as that gives the number of ways in which we can select 4, 3, 2...so that also makes life a little bit easier... regards scrabbler
