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18 Sep 2024, 01:30
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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43% (02:56) correct
57%
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wrong
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N is called a sweet number if the sum of its digits is an even number. How many four-digit numbers, with distinct digits, are sweet numbers?
A. 2160
B. 2376
C. 3672
D. 4200
E. 4500
A. 2160
B. 2376
C. 3672
D. 4200
E. 4500
18 Sep 2024, 02:06
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Sum of digits of four-digit number \(abcd\) is even when:
Case 1: All even digits - \(0, 2, 4, 6, 8\)
There are \(4\) ways to select an even number for \(a\), since it can not be \(0\). Out of remaining \(4\), \(3\) digits for \(b,c,d\) can be selected in \(4C3=4\) ways and arranged in \(3!=6\) ways
\(\text{Case 1}=4*4*6=96\)
Case 2: Two even (\(0, 2, 4, 6, 8\)) & two odd (\(1,3,5,7,9\)) digits, Starting with an even number
There are \(4\) ways to select an even number for \(a\), since it can not be \(0\). Remaining even number can be selected in \(4\) ways. Two odd numbers out of \(5\) in \(5C2=10\) ways. Two odd and one even number can be arranged in \(3!=6\) ways
\(\text{Case 2}=4*4*10*6=960\)
Case 3: Two even & two odd digits, Starting with an odd number
First odd digit has \(5\) ways. Second off digit has \(4\) ways. Two even numbers out of \(5\) in \(5C2=10\) ways. Two even and one odd number can be arranged in \(3!=6\) ways
\(\text{Case 3}=5*4*10*6=1200\)
Case 4: All odd
4 distinct digits out of 5 odd numbers can be selected & arranged in \(5*4*3*2=120\) ways
\(\text{Total four-digit sweet numbers}=96+960+1200+120=2376\)
Answer: B
Case 1: All even digits - \(0, 2, 4, 6, 8\)
There are \(4\) ways to select an even number for \(a\), since it can not be \(0\). Out of remaining \(4\), \(3\) digits for \(b,c,d\) can be selected in \(4C3=4\) ways and arranged in \(3!=6\) ways
\(\text{Case 1}=4*4*6=96\)
Case 2: Two even (\(0, 2, 4, 6, 8\)) & two odd (\(1,3,5,7,9\)) digits, Starting with an even number
There are \(4\) ways to select an even number for \(a\), since it can not be \(0\). Remaining even number can be selected in \(4\) ways. Two odd numbers out of \(5\) in \(5C2=10\) ways. Two odd and one even number can be arranged in \(3!=6\) ways
\(\text{Case 2}=4*4*10*6=960\)
Case 3: Two even & two odd digits, Starting with an odd number
First odd digit has \(5\) ways. Second off digit has \(4\) ways. Two even numbers out of \(5\) in \(5C2=10\) ways. Two even and one odd number can be arranged in \(3!=6\) ways
\(\text{Case 3}=5*4*10*6=1200\)
Case 4: All odd
4 distinct digits out of 5 odd numbers can be selected & arranged in \(5*4*3*2=120\) ways
\(\text{Total four-digit sweet numbers}=96+960+1200+120=2376\)
Answer: B
General Discussion
18 Sep 2024, 02:58
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tgsankar10
Sum of digits of four-digit number \(abcd\) is even when:
Case 1: All even digits - \(0, 2, 4, 6, 8\)
There are \(4\) ways to select an even number for \(a\), since it can not be \(0\). Out of remaining \(4\), \(3\) digits for \(b,c,d\) can be selected in \(4C3=4\) ways and arranged in \(3!=6\) ways
\(\text{Case 1}=4*4*6=96\)
Case 2: Two even (\(0, 2, 4, 6, 8\)) & two odd (\(1,3,5,7,9\)) digits, Starting with an even number
There are \(4\) ways to select an even number for \(a\), since it can not be \(0\). Remaining even number can be selected in \(4\) ways. Two odd numbers out of \(5\) in \(5C2=10\) ways. Two odd and one even number can be arranged in \(3!=6\) ways
\(\text{Case 2}=4*4*10*6=960\)
Case 3: Two even & two odd digits, Starting with an odd number
First odd digit has \(5\) ways. Second off digit has \(4\) ways. Two even numbers out of \(5\) in \(5C2=10\) ways. Two even and one odd number can be arranged in \(3!=6\) ways
\(\text{Case 3}=5*4*10*6=1200\)
Case 4: All odd
4 distinct digits out of 5 odd numbers can be selected & arranged in \(5*4*3*2=120\) ways
\(\text{Total four-digit sweet numbers}=96+960+1200+120=2376\)
Answer: B
Case 1: All even digits - \(0, 2, 4, 6, 8\)
There are \(4\) ways to select an even number for \(a\), since it can not be \(0\). Out of remaining \(4\), \(3\) digits for \(b,c,d\) can be selected in \(4C3=4\) ways and arranged in \(3!=6\) ways
\(\text{Case 1}=4*4*6=96\)
Case 2: Two even (\(0, 2, 4, 6, 8\)) & two odd (\(1,3,5,7,9\)) digits, Starting with an even number
There are \(4\) ways to select an even number for \(a\), since it can not be \(0\). Remaining even number can be selected in \(4\) ways. Two odd numbers out of \(5\) in \(5C2=10\) ways. Two odd and one even number can be arranged in \(3!=6\) ways
\(\text{Case 2}=4*4*10*6=960\)
Case 3: Two even & two odd digits, Starting with an odd number
First odd digit has \(5\) ways. Second off digit has \(4\) ways. Two even numbers out of \(5\) in \(5C2=10\) ways. Two even and one odd number can be arranged in \(3!=6\) ways
\(\text{Case 3}=5*4*10*6=1200\)
Case 4: All odd
4 distinct digits out of 5 odd numbers can be selected & arranged in \(5*4*3*2=120\) ways
\(\text{Total four-digit sweet numbers}=96+960+1200+120=2376\)
Answer: B
Case 2: Two even () & two odd () digits, Starting with an even number
Can you help me understand why the following is wrong for this scenario-
1st digit can have 4 possibilities. Second even has 4 possibilities. The odd numbers have 5 and 4 possibilities.
Hence 4*4*5*4
Combinations of the last 3 digits - 3!
Hence- 4*4*5*4*3!
I seem to be counting twice of what you are counting, but I do not understand why this method is wrong.
Posted from my mobile device
