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15 Jun 2021, 21:56
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
56% (01:37) correct
44%
(02:05)
wrong
based on 95
sessions
History
Date
Time
Result
Not Attempted Yet
In how many ways can peter select 3 cards from a card pack such that all three belong to different suits?
A.2197
B.8112
C.8788
D.17576
E.28561
A.2197
B.8112
C.8788
D.17576
E.28561
15 Jun 2021, 22:11
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Yogananda
In how many ways can peter select 3 cards from a card pack such that all three belong to different suits?
A.2197
B.8112
C.8788
D.17576
E.28561
A.2197
B.8112
C.8788
D.17576
E.28561
There are 4 different suits in a card pack. So, ways to choose 3 out of these = 4C3=4.
Each suit has 13 cards, so we can choose any of the 13 cards. Ways to choose from 3 different suits = 13*13*13
Total ways from the card pack = 4*13*13*13=8788
C
16 Jun 2021, 00:19
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Solution:
In a standard pack of cards there are :
The question asks us the number of ways of selecting 3 cards of 3 different suits.
First of all, we have to choose the 3 suits from 4 suits available, which can be done in \(4C3 = 4\) ways.
After choosing 3 suits the number of ways of choosing a card from each = \(13 \times 13 \times 13\)
Thus the number of ways of selecting 3 cards of 3 different suits = \(4C3 \times 13 \times 13 \times 13 = 4 \times 13 \times 13 \times 13 = 8788\)
Hence the right answer is Option C.
In a standard pack of cards there are :
- 1. 4 suits: Heart, Club, Diamond, Spade
- 2. Each suit has 13 cards
The question asks us the number of ways of selecting 3 cards of 3 different suits.
First of all, we have to choose the 3 suits from 4 suits available, which can be done in \(4C3 = 4\) ways.
After choosing 3 suits the number of ways of choosing a card from each = \(13 \times 13 \times 13\)
Thus the number of ways of selecting 3 cards of 3 different suits = \(4C3 \times 13 \times 13 \times 13 = 4 \times 13 \times 13 \times 13 = 8788\)
Hence the right answer is Option C.
