Combinations is a mathematical procedure to compute the number of ways in which k items can be selected from n different items without replacement and irrespective of the order.

${n\choose k}=\frac{n!}{k!(n-k)!}$

Permutation is a mathematical procedure to determine the number of arrangements of k items from n different items respective of the order of arrangement.

$^{n}P_{k}=\frac{n!}{(n-k)!}$

In this case we need to select two different cards from a pack of 52 cards.

Two cards are selected without replacement:

Compute the number of ways to select 2 cards from 52 cards without replacement as follows:

${n\choose k}=\frac{n!}{k!(n-k)!}$

${52\choose 2}=\frac{52!}{2!(52-2)!}$

$=\frac{52\times 51\times 50!}{2!\times50!}\\=1326$

Thus, the number of ways to select 2 cards from 52 cards without replacement is 1326.

Two cards are selected and the order matters.

Compute the number of ways to select 2 cards from 52 cards in case the order is important as follows:

$^{n}P_{k}=\frac{n!}{(n-k)!}$

$^{52}P_{2}=\frac{52!}{(52-2)!}$

$=\frac{52\times 51\times 52!}{50!}$

$=52\times 51\\=2652$

Thus, the number of ways to select 2 cards from 52 cards in case the order is important is 2652.

6 0

3 years ago

Five plus twice a number is thirty-one

igomit [66]

5 + 2n = 31
2n = 26
n = 13

3 0

3 years ago

Read 2 more answers

A scientist inoculates mice, one at a time, with a disease germ until he finds 2 that have contracted the disease. if the probab

AnnyKZ [126]

There are several possible events that lead to the eighth mouse tested being the second mouse poisoned. There must be only a single mouse poisoned before the eighth is tested, but this first poisoning could occur with the first, second, third, fourth, fifth, sixth, or seventh mouse. Thus there are seven events that describe the scenario we are concerned with. With each event, we want two particular mice to become diseased (1/6 chance) and the remaining six mice to remain undiseased (5/6 chance). Thus, for each of the seven events, the probability of this event occurring among all events is (1/6)^2(5/6)^6. Since there are seven of these events which are mutually exclusive, we sum the probabilities: our desired probability is 7(1/6)^2(5/6)^6 = (7*5^6)/(6^8).

5 0

3 years ago