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805+ (Hard)|   Combinations|      
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N is called a sweet number if the sum of its digits is an even number. 18 Sep 2024, 01:30
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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


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B
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tgsankar10
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N is called a sweet number if the sum of its digits is an even number. 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
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N is called a sweet number if the sum of its digits is an even number. 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
Show more

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.

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tgsankar10
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N is called a sweet number if the sum of its digits is an even number. 18 Sep 2024, 03:21
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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 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
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"1st digit can have 4 possibilities. Second even has 4 possibilities. The odd numbers have 5 and 4 possibilities."

Here the two odd numbers are already arranged in 5*4=20 ways. The remaining even number has to be arranged in 3 ways.

For example: in \(bcd\), c & d are odd numbers. 5 possibilities for c & 4 for d. b can be placed in 3 positions such as: bcd, cbd, cdb. Then we will get 4*4*5*4*3=960

But in your case the permutations for two odd numbers are performed twice. That is the reason for double counting.
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N is called a sweet number if the sum of its digits is an even number. 18 Sep 2024, 11:32
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The cases how we can get the "sweet numbers" are
  1. All Odds
  2. All Evens
  3. Two Evens and Two Odds

Case (1)
For all odds, the total arrangements that we have = \(5C4*4! = 120\)

Case (2)
For all Evens, the total arrangements that we have = All cases - The cases where zero comes first = \( (5C4*4!) - (1*4C3*3!) = 96\)

Case (3)
For all Evens, the total arrangements that we have
= All cases - The cases where zero comes first
= \((5C2*5C2*4!)-(1*4C1*5C2*3!)\)
\(\\
= 2400-240\\
= 2160\\
\)

So, Total Cases = 120+96+2160 = 2376. OPTION B
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N is called a sweet number if the sum of its digits is an even number. 18 Sep 2024, 12:17
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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 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

"1st digit can have 4 possibilities. Second even has 4 possibilities. The odd numbers have 5 and 4 possibilities."

Here the two odd numbers are already arranged in 5*4=20 ways. The remaining even number has to be arranged in 3 ways.

For example: in \(bcd\), c & d are odd numbers. 5 possibilities for c & 4 for d. b can be placed in 3 positions such as: bcd, cbd, cdb. Then we will get 4*4*5*4*3=960

But in your case the permutations for two odd numbers are performed twice. That is the reason for double counting.
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I don't quite understand the explanation.
In bcd -
Why are we not considering
dcb, dbc and bdc as valid options ?
All digits are distinct, so all these combinations will result in separate numbers. The only case we've marked is that the first number is even/odd. Why are we performing the operation 3!/2! and not just multiplying with 3! ?
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N is called a sweet number if the sum of its digits is an even number. 18 Sep 2024, 12:53
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tgsankar10, nvm, I think I understand this now. Is there any study material that can be referenced specifically for probability and P&C (preferably on gmatclub).
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N is called a sweet number if the sum of its digits is an even number. 18 Sep 2024, 22:50
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Bunuel
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
Show more
Interesting question!
N is 4 digit sweet number whose sum of its digit is an even number with distinct digits.
When that will be possible.
1st when all are even. e+e+e+e=even
2nd when all are odd. o+o+o+o = even
3rd when 2 even and 2 odd = e+o+e+o..... (total of 6 cases) (eeoo, eoeo, eooe, ooee, oeoe, oeeo) you can count this with factorial method as well \(\frac{4!}{2!2!}\)
you see in 3rd case there are 3 cases which are starting with even and 3 are starting with odd.
now lets start calculating.

1st - when all are even. e+e+e+e=even
_ + _ + _ + _ = Even
remember here at first digit place "0" is not possible.
so out of those 5 (0,2,4,6,8) at first place 4 (2,4,6,8) digits are possible.
Total cases = 4*4*3*2 = 96

2nd - 2nd when all are odd. o+o+o+o = even
Here, Total cases = 5*4*3*2 = 120

3rd when 2 even and 2 odd = e+o+e+o..... (total of 6 cases)
i - 3 cases which are starting with even
Total cases = 3*(4*4*5*4) = 960

ii - 3 are starting with odd.
Total cases = 3*(5*4*5*4) = 1200

Therefore total cases combining all possible counts = 96+120+960+1200 = 2376
Answer is B.
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