1. Four cards are drawn in succession, without replacement, from a standard deck of 52 cards. How many sets of four cards are possible?
270,725
6,497,400
54,145
1,082,900
2. A cola company decides to test five different brands of soft drink. The different brands have been labeled F, G, H, I, and J. The company decides to compare each brand with the other brands by pairing them together. How many different pairs will result from selecting two different brands at a time?
25
20
12
10
Looks like both questions deal with combination, that is, the order
of selecting objects does not matter.
#1 , you are choosing 4 from 52 which is
C(52,4) or 52C4
= 270725
most scientific calculators have that as a function.
look for nCr, on mine it is paired with the "5"
enter:
52
2ndF
4
#2, how many groups of 2 can you choose from 5 ?
=
i still don't understand number two though
1. To find the number of sets of four cards that can be drawn from a standard deck of 52 cards, we can use the concept of combinations.
The formula for combinations is given by nCr = n! / (r!(n-r)!), where n is the total number of items and r is the number of items being selected without replacement.
In this case, we have n = 52 (total number of cards in the deck) and r = 4 (number of cards being drawn without replacement).
So, the number of sets of four cards possible is calculated as 52C4 = 52! / (4!(52-4)!) = 52! / (4!48!) = (52 * 51 * 50 * 49) / (4 * 3 * 2 * 1) = 270,725.
Therefore, the correct answer is option 1: 270,725.
2. To find the number of different pairs that can be formed by selecting two brands at a time from five different brands, we can use the concept of combinations again.
Using the same formula as before, nCr = n! / (r!(n-r)!), we have n = 5 (total number of brands) and r = 2 (number of brands being selected without replacement).
So, the number of different pairs possible is calculated as 5C2 = 5! / (2!(5-2)!) = 5! / (2!3!) = (5 * 4) / (2 * 1) = 10.
Therefore, the correct answer is option 4: 10.
Thanks though, I passed the quiz
#2 each of the 5 can pair with any of the other 4 ... 5 * 4 = 20
... BUT ... A and B is the same group as B and A ... so, only half of 20