Len H. answered • 04/22/24
Mathematics Literacy Specialist
First thing to remember is the difference between combinations and permutations. In combinations, the order does not matter (XY is same as YX). In permutations, the order is important (XY and YX are two separate things).
Next, you can list out all possibilities and count them OR you can use the particular formulas.
nCr = (n!)/[((n-r)!)(r!)]
nPr = (n!)/[(n-r)!]
Len H.
n is the total choices possible. r is the number of choices you want. Think of 5C3 as 5 choices and you are selecting 3 of them with no order.04/22/24
Len H.
Looking at the way it is written in your problem, n is the top number and r is the bottom number. So it will be 5C3 x 10C204/22/24
Charles T.
Noted. Thank you.04/22/24
Len H.
Please share your answers so we can check04/23/24
Charles T.
Right so for a) I ended up with 10 × 45 = 450, and for b) my working out stalled at ((n-4)!)/((n-6)!) = (11)(n!)...04/23/24
Charles T.
Hi Sir. Thanks for this explanation. So in that case is it read as: a) 5C3 × 10C2 (or 3C5) × 2C10) and b) is nP6 = nP4 × 4P2 ??04/22/24