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VeritasKarishma
kchen1994
How many organizational structures can be formed that consist of Division A; at least one of Divisions B and C; at least two of Divisions D, E, and F; and at least two of Divisions G, H, J, and K?

a) 60
b) 90
c) 99
d) 120
e) 132

Division A can be taken in 1 way.

Ways of selecting at least one of B and C = 2^2 - 1 = 3
Explanation: B can be picked in 2 ways (pick or not pick) and C can be picked in 2 ways (pick or not pick). This gives us 2*2 = 4 total ways. Out of this, it is not acceptable to not pick either so we remove 1.

Ways of selecting at least two of D, E and F = 2^3 - 1 - 3 = 4
Explanation: D can be picked in 2 ways (pick or not pick), E can be picked in 2 ways (pick or not pick) and F can be picked in 2 ways (pick or not pick). This gives us 2*2*2 = 8 total ways. Out of this, it is not acceptable to not pick any so we remove 1. It is also not acceptable to pick only 1 (since at least 2 have to picked) so remove 3 ways in which you can pick any one.

Ways of selecting at least 2 of G, H, J, K = 2^4 - 1 - 4 = 11
Same explanation as above.

Total = 1*3*4*11 = 132


Understood. Thanks for a quick response VeritasKarishma
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kchen1994
How many organizational structures can be formed that consist of Division A; at least one of Divisions B and C; at least two of Divisions D, E, and F; and at least two of Divisions G, H, J, and K?

a) 60
b) 90
c) 99
d) 120
e) 132

Hi ParthSanghavi,

The way would be to see how each set will turn out to be..
consist of Division A - 1 way;
at least one of Divisions B and C - there will be one way when none will be there, so Total-1=2^2-1=3;
at least two of Divisions D, E, and F- none is there-1 way and one is there-3 ways, so total-(1+3)=2^3-4; and
at least two of Divisions G, H, J, and K- none is there-1 way and one is there-4 ways, so total-(1+3)=2^4-5;

total ways = \(1*3*(2^3-4)(2^4-5)=1*3*4*11=132\)
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chetan2u
kchen1994
How many organizational structures can be formed that consist of Division A; at least one of Divisions B and C; at least two of Divisions D, E, and F; and at least two of Divisions G, H, J, and K?

a) 60
b) 90
c) 99
d) 120
e) 132

Hi ParthSanghavi,

The way would be to see how each set will turn out to be..
consist of Division A - 1 way;
at least one of Divisions B and C - there will be one way when none will be there, so Total-1=2^2-1=3;
at least two of Divisions D, E, and F- none is there-1 way and one is there-3 ways, so total-(1+3)=2^3-4; and
at least two of Divisions G, H, J, and K- none is there-1 way and one is there-4 ways, so total-(1+3)=2^4-5;

total ways = \(1*3*(2^3-4)(2^4-5)=1*3*4*11=132\)

Seems, Karishma has also replied in the meantime. Thanks
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hi experts,

could anybody help me in explaining the question please, i am struck in understanding the question per se...

thanks
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push12345
For this question one needs to manually calculate cases

1*(2C1+2C2)*(3C2+3C3)*(4C2+4C3+4C4)
=1*3*4*11
=132

Posted from my mobile device

One also does not quote simply C1/C2 to make a problem more complex.
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ccheryn
hi experts,

could anybody help me in explaining the question please, i am struck in understanding the question per se...

thanks

I have the same question ....
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