Triple Combination
Calculate the combinations: 8C4 x 6C3 x 4C2
Expand Hint
$$$C_{(n,r)}=\frac{P_{(n,r)}}{r!}=\frac{n!}{[r!(n-r)!]}$$$
where
$$C_{n,r}$$
is the number of different combinations of
$$n$$
distinct objects taken
$$r$$
at a time, and
$$P$$
is the number of different permutations.
Hint 2
Combination format:
$$nCr$$
$$$C_{(n,r)}=\frac{P_{(n,r)}}{r!}=\frac{n!}{[r!(n-r)!]}$$$
where
$$C_{n,r}$$
is the number of different combinations of
$$n$$
distinct objects taken
$$r$$
at a time, and
$$P$$
is the number of different permutations.
Solving the first combination:
$$$C_{(8,4)}=\frac{8!}{4![(8-4)!]}=\frac{8!}{4!(4!)}=\frac{8\cdot 7\cdot 6\cdot 5\cdot 4!}{4\cdot 3\cdot 2\cdot 1\cdot (4!)}=\frac{1,680}{24}=70$$$
Solving the second combination:
$$$P_{(6,3)}=\frac{6!}{3![(6-3)!]}=\frac{6!}{3!(3!)}=\frac{6\cdot 5\cdot 4\cdot 3!}{3\cdot 2\cdot 1\cdot 3!}=\frac{120}{6}=20$$$
Solving the third combination:
$$$P_{(4,2)}=\frac{4!}{2![(4-2)!]}=\frac{4!}{2!(2!)}=\frac{4\cdot 3 \cdot 2!}{2\cdot 1\cdot 2!}=\frac{12}{2}=6$$$
Multiplying the three combinations together:
$$$70 \times 20 \times 6=8,400$$$
8,400
Time Analysis
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