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How many 4-digit positive integers are there in which half of the digi

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How many 4-digit positive integers are there in which half of the digi  [#permalink]

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New post 28 Oct 2018, 23:40
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How many 4-digit positive integers are there in which half of the digits are even and another half of the digit is odd?

A. 625
B. 1,125
C. 1,250
D. 9,000
E. 10,000

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Re: How many 4-digit positive integers are there in which half of the digi  [#permalink]

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New post 29 Oct 2018, 02:25
Total no. of ways in which out of 4 digits 2 are even and 2 are odd is \(\frac{4!}{2! * 2!}\) = 6 ways

The 6 cases are EEOO, EOEO, EOOE, OOEE, OEOE, OEEO. In these 3 cases start with Even numbers and 3 with odd numbers. Total no. of even numbers 0,2,4,6,8. Odd no.'s are 1,3,5,7,9.

1) EEOO = 4*5*5*5 = 500 numbers --> Since out of 5 even numbers 0 cant take first place so a total of 4 ways to fill that place and remaining each can be filled in 5 ways (as repetition is allowed)

2) EOEO = 4*5*5*5 = 500 numbers

3) EOOE = 4*5*5*5 = 500 numbers

4) OOEE = 5*5*5*5 = 625 numbers --> all places can be filled in 5 ways

5) OEOE = 5*5*5*5 = 625 numbers

6) OEEO = 5*5*5*5 = 625 numbers

Finally, 500*3 + 625*3 = 1500 + 1875 = 3375 numbers. I'm getting this answer. Is there any mistake in my approach? What is the correct answer and how?
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How many 4-digit positive integers are there in which half of the digi  [#permalink]

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New post 29 Oct 2018, 04:31
Two E and Two O
Since the first digit can't be zero, let us first fix that.
Case 1 E is First digit Any two digits O and last digit is E
No of ways = 4*5*5*5
Case 2 O is first digit, any two digits are E and last one is O
No of ways = 5*5*5*5
Case 1 + case 2 = 1125
.....

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Re: How many 4-digit positive integers are there in which half of the digi  [#permalink]

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New post 29 Oct 2018, 04:36
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Bunuel wrote:
How many 4-digit positive integers are there in which half of the digits are even and another half of the digit is odd?

A. 625
B. 1,125
C. 1,250
D. 9,000
E. 10,000


Case 1: When Thousands place is Even
Thousands place can be filled in 4 ways using digits 2, 4, 6, 8
One more place for even digit can be chosen in 3 ways and may be filled in 5 ways using digits 0, 2, 4, 6, 8
other two places may be filled in 5*5 ways using digits 1, 3, 5, 7, 9
i.e. Total Outcomes = 4*3*5*5*5 = 1500

Case 2: When Thousands place is ODD
Thousands place can be filled in 5 ways using digits 1, 3, 5, 7, 9
two places for even digits may be chosen in 3C2 = 3 ways and may be filled in 5*5 ways using digits 0, 2, 4, 6, 8
remaining place may be filled in 5 ways using digits 1, 3, 5, 7, 9
i.e. Total Outcomes = 5*3*5*5*5 = 1875

Total Such numbers = 1500+1875 = 3375

Bunuel Answer Options should be adjusted accordingly...
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How many 4-digit positive integers are there in which half of the digi  [#permalink]

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New post Updated on: 29 Oct 2018, 04:47
GMATinsight Thank you. Even I was surprised as to how my answer was correct when I knowingly didn't arrange the digits.
I'm still happy that my approach of fixing first position was correct :p

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Originally posted by ShankSouljaBoi on 29 Oct 2018, 04:46.
Last edited by ShankSouljaBoi on 29 Oct 2018, 04:47, edited 1 time in total.
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How many 4-digit positive integers are there in which half of the digi &nbs [#permalink] 29 Oct 2018, 04:46
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