(i) if at least 5 must be from the same suit then the soln is in two cases:lyounamu said:Another question guys:
A standard pack of 52 cards consists of 13 cards of ecah of the four suits: spades, hearts, clubs and diamonds:
i) In how many ways can six cards be selected without replacement if at least fiv must be of the same suit? (order is not important)
ii) In how many ways can six cards be selected without replacement so that exactly two are spades and four are clubs? (order is NOT improtant)
Come on guys~
i) 6 cards:samwell said:(i) if at least 5 must be from the same suit then the soln is in two cases:
5 cards from same suit selected[13C5*39C1]
+6 cards from same suit[13C6*39C0]
(ii)13C2*13C4*26C0
is this right????? not sure how to treat the 2nd part
i) 13C6 x 4 is for each suit given that all six cards selected are the same suit. 13C6 is selecting 6 from 13 (one suit) and since there are 4 suits you multiply by four.lyounamu said:Another question guys:
A standard pack of 52 cards consists of 13 cards of ecah of the four suits: spades, hearts, clubs and diamonds:
i) In how many ways can six cards be selected without replacement if at least fiv must be of the same suit? (order is not important)
ii) In how many ways can six cards be selected without replacement so that exactly two are spades and four are clubs? (order is NOT improtant)
Come on guys~
Couldnt' have put it better myself. Well done, bored of sc. Very concisely and well explained.bored of sc said:i) 13C6 x 4 is for each suit given that all six cards selected are the same suit. 13C6 is selecting 6 from 13 (one suit) and since there are 4 suits you multiply by four.
13C5 x 39C1 x 4 is for each suit given that five of one suit is selected. 13C5 is selecting 5 of the one suit from the 13 of one suit. 39C1 is choosing 1 from the remaining 3 suits (can't be the same suit as the other five cause that would make is 13C6) which is 13 (number of cards in one suit) multiplied by 3 - remember to multiply this answer by 4 since any suit can have the 5 selections.
Since you can have 6 of one suit OR 5 of one suit and any other card you ADD the values you get for each case.
13C6*4 + 13C5*39C1*4 = value Namu said (I hope).
I just read over that and realised it was the most confusing piece of explanation ever written. Sorry about that.
In case it is incorrect, the answer I got was 207636. You said it was 200772.lyounamu said:Couldnt' have put it better myself. Well done, bored of sc. Very concisely and well explained.
i) A = root (22 + rt(12)2)midifile said:i) Express rt(12)sinx + 2 cosx in the form Acos(x-@)
ii) State the number of solutions that satisfy the equation rt(12)sinx + 2 cosx = 1 in the domain 0<=x<=2pi
iii) Write down the general solution to rt(12)sinx + 2 cosx = 1
uh wait. Let me see:bored of sc said:In case it is incorrect, the answer I got was 207636. You said it was 200772.
Yay!lyounamu said:uh wait. Let me see:
I miscalculated it. Your one is right.
Thank you so much. I will have a look at that more closelyQuLiT said:for your question namu:
il try and make it as clear as possible but yaeh not so good with typing
using cartesian equation:
let y = 0 x = R+r
0 = h + (r+R)tan@ - g(r+R)^2/2v^2 cos@
you use v^2 from part c then there for
0> h+ (r+R)tan@ -((r+R)^2 tan @ )/R
then after this you factorise the tan@ and rearrange and you should get your answer if your still having troubles tell me and il scan my working