A bag contains 5 green and 7 red jellybeans. How many ways can 5 jellybeans be withdrawn from the bag so that the number of green ones withdrawn will be less than 2?
We use the formula:
\[ C(n, k) = \frac{n!}{k!(n-k)!} \]
For this case, \(n = 7\) (total red jellybeans) and \(k = 5\) (jellybeans to be withdrawn), so:
\[ C(7, 5) = \frac{7!}{5!(7-5)!} \]
If 1 green jellybean is withdrawn, 4 red jellybeans must also be withdrawn. The number of ways to do this is the product of the combinations of choosing 1 green from 5 and 4 red from 7:
\[ C(5, 1) \times C(7, 4) \]
\[ C(5, 1) = \frac{5!}{1!(5-1)!} \]
\[ C(7, 4) = \frac{7!}{4!(7-4)!} \]
Summing up the above:
\[ \text{Total number of ways} = C(7, 5) + (C(5, 1) \times C(7, 4)) \]
Which gives us gives us:
\[ 196 \]
Answer: 196 ways
A certain congressional committee consists of 14 senators and 13 representatives. How many ways can a subcommittee of 5 be formed if at least 4 of the members must be representatives?
\[ C(13, 5) = \frac{13!}{5!(13-5)!} \]
\[ C(13, 4) \times C(14, 1) = \frac{13!}{4!(13-4)!} \times \frac{14!}{1!(14-1)!} \]
Summing the two expressions above:
\[ \text{Total number of ways} = C(13, 5) + (C(13, 4) \times C(14, 1)) \]
We get:
\[ 11297 \]
Answer: 11,297 ways
If a coin is tossed 5 times, and then a standard six-sided die is rolled 2 times, and finally a group of three cards are drawn from a standard deck of 52 cards without replacement, how many different outcomes are possible?
We have to multiply the number of outcomes from each event:
When the coin is tossed 5 times, the number of outcomes is \(2^5\).
If the die is rolled 2 times, the number of outcomes is \(6^2\).
We can use the combination formula for the cards since the order in which the cards are drawn does not matter:
\[ C(52, 3) = \frac{52!}{3!(52-3)!} \]
Thus, the total number of different outcomes for the entire sequence of events is the product of the outcomes for each event:
\[ \text{Total outcomes} = 2^5 \times 6^2 \times C(52, 3) \]
Which gives us:
\[25459200 \]
Answer: 25,459,200