easy question (1 Viewer)

[ToO LaZy ^*]

the staff in an office consists of 4 males and 7 females
how many committees of 5 staff can be chosen which contain exactly 3 females

 

[gordo]

11c5 (7/11)^3. (4/11)^2

 

[CrashOveride]

7C3 x 4C2

 

[gordo]

oops i did probability,
sorry

 

[ToO LaZy ^*]

excellent..

 

[ToO LaZy ^*]

another question...

prove by induction that, for all n=>1
(n+1)(n+2)...(2n-1)2n = 2ⁿ[1x3x...x(2n-1)]

 

[CrashOveride]

11c5 (7/11)^3. (4/11)^2

So it's about 15.74? hmm i get 5/11

 

[ToO LaZy ^*]

another question...

prove by induction that, for all n=>1
(n+1)(n+2)...(2n-1)2n = 2ⁿ[1x3x...x(2n-1)]

help?...anyone?

 

[withoutaface]

Ill just do it for k and k+1

(k+1)(k+2)...(2k-1)2k=2^k[1x3x...x(2k-1)]

RTP

LHS=(k+2)...(2k-1)(2k)(2k+1)(2k+2)=2^k[1x3x...x(2k-1)](2k)(2k+1)(2k+2)/(k+1)=2^k+1[1x3x...x(2k-1)x(2k)x(2k+1)]=RHS
this is the original part then we sub in the value from S(k) but we must divide by k+1 then the(2k+2) makes 2^k--->2^k+1 and cancels the /k+1

 

[ToO LaZy ^*]

thx champ!