(a) If 13 cards are randomly selected from a standard 52-card deck, must at least 2 be of the same denomination (2, 3, 4, ☐ ☐ 00 ..., J, Q, K, A)? Why? (Select all that apply.) Yes. For example, six hearts: 2, 3, 4, 5, 6, 7, and seven diamonds: 2, 3, 4, 5, 6, 7, 8, could be selected. Six of these are of the same denomination. No. For example, thirteen hearts could be selected: 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, A. No two of these are of the same denomination. No. For example, the 2, 4, 6 of hearts, the 5, 10, K of diamonds, the 8, 9, J, A of clubs, and the 3, 7, Q of spades have no two cards that are of the same denomination. No. For example, thirteen hearts could be selected: 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, A. At least two of these are of the same denomination. No. For example, six hearts: 2, 3, 4, 5, 6, 7, and seven diamonds: 2, 3, 4, 5, 6, 7, 8, could be selected. Six of these are of the same denomination. × (b) If 20 cards are randomly selected from a standard 52-card deck, must at least 2 be of the same denomination (2, 3, 4, .*** J, Q, K, A)? Why? Yes . Let A be the set of 20 cards selected from the 52-card deck, and let B be the 13 different denominations of cards in the deck. If we construct a function from A to B, then by the pigeonhole not a one-to-one correspondence ☑☑. Therefore, it is impossible ☑× to randomly select 20 cards with no repeated denomination from a standard 52-card deck. principle, the function must be
(a) If 13 cards are randomly selected from a standard 52-card deck, must at least 2 be of the same denomination (2, 3, 4, ☐ ☐ 00 ..., J, Q, K, A)? Why? (Select all that apply.) Yes. For example, six hearts: 2, 3, 4, 5, 6, 7, and seven diamonds: 2, 3, 4, 5, 6, 7, 8, could be selected. Six of these are of the same denomination. No. For example, thirteen hearts could be selected: 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, A. No two of these are of the same denomination. No. For example, the 2, 4, 6 of hearts, the 5, 10, K of diamonds, the 8, 9, J, A of clubs, and the 3, 7, Q of spades have no two cards that are of the same denomination. No. For example, thirteen hearts could be selected: 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, A. At least two of these are of the same denomination. No. For example, six hearts: 2, 3, 4, 5, 6, 7, and seven diamonds: 2, 3, 4, 5, 6, 7, 8, could be selected. Six of these are of the same denomination. × (b) If 20 cards are randomly selected from a standard 52-card deck, must at least 2 be of the same denomination (2, 3, 4, .*** J, Q, K, A)? Why? Yes . Let A be the set of 20 cards selected from the 52-card deck, and let B be the 13 different denominations of cards in the deck. If we construct a function from A to B, then by the pigeonhole not a one-to-one correspondence ☑☑. Therefore, it is impossible ☑× to randomly select 20 cards with no repeated denomination from a standard 52-card deck. principle, the function must be
College Algebra (MindTap Course List)
12th Edition
ISBN:9781305652231
Author:R. David Gustafson, Jeff Hughes
Publisher:R. David Gustafson, Jeff Hughes
Chapter8: Sequences, Series, And Probability
Section8.6: Permutations And Combinations
Problem 4SC
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Transcribed Image Text:(a) If 13 cards are randomly selected from a standard 52-card deck, must at least 2 be of the same denomination (2, 3, 4,
☐ ☐
00
..., J, Q, K, A)? Why? (Select all that apply.)
Yes. For example, six hearts: 2, 3, 4, 5, 6, 7, and seven diamonds: 2, 3, 4, 5, 6, 7, 8, could be selected. Six of these are of the same denomination.
No. For example, thirteen hearts could be selected: 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, A. No two of these are of the same denomination.
No. For example, the 2, 4, 6 of hearts, the 5, 10, K of diamonds, the 8, 9, J, A of clubs, and the 3, 7, Q of spades have no two cards that are of the same
denomination.
No. For example, thirteen hearts could be selected: 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, A. At least two of these are of the same denomination.
No. For example, six hearts: 2, 3, 4, 5, 6, 7, and seven diamonds: 2, 3, 4, 5, 6, 7, 8, could be selected. Six of these are of the same denomination.
×
(b) If 20 cards are randomly selected from a standard 52-card deck, must at least 2 be of the same denomination (2, 3, 4,
.***
J, Q, K, A)? Why?
Yes
. Let A be the set of 20 cards selected from the 52-card deck, and let B be the 13
different denominations of cards in the deck. If we construct a function from A to B, then by the pigeonhole
not a one-to-one correspondence ☑☑. Therefore, it is impossible ☑× to randomly select 20 cards with no repeated denomination from a standard 52-card deck.
principle, the function must be
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