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Re: How many five-digit even numbers can be formed using
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03 Oct 2019, 14:29
1
Without restrictions, there can be 5x4x3x2x1=120 combinations of 5 numbers. However, we want 5 digit even numbers here, so 1 and 3 cannot be the ending bit, and 0 cannot be the starting bit. So the final answer will be 120-(numbers starting with 0)-(odd numbers)+(odd numbers starting with 0) 1/5 of these 120 numbers will start with a 0, so 120/5=24 2/5 of these 120 numbers will end with 1 or 3, so 120 x 2/5=48 For odd numbers starting with 0, because 0 cannot be the ending bit in this situation, so only half of these 24 numbers starting with 0 will be odd numbers. 24/2=12 The answer will be 120-24-48+12=60 D. Literally the hardest GRE math problem I have seen
Since the resulting number must be EVEN, let's consider two possible cases:
case i: The unit's digit is 0 case ii: The unit's digit is 2 or 6
Question: Why am I considering those two cases separately? Answer: I'm doing so because we DON'T get a 5-digit number if the first digit is 0 (e.g., 02631 is not a 5-digit number)
case i: The unit's digit (aka the 5th digit) is 0 So, there is 1 way to assign a value to the 5th digit. There are now 4 digits remaining to choose from. So, we can assign a value to the 1st digit in 4 ways. We can assign a value to the 2nd digit in 3 ways. We can assign a value to the 3rd digit in 2 ways. We can assign a value to the 4th digit in 1 way. By the Fundamental Counting Principal (FCP), we can complete all five steps (and create a 5 digit number) in (1)(4)(3)(2)(1) ways (= 24 ways)
case ii: The unit's digit is 2 or 6 Since the last digit is 2 or 6, there are 2 ways to assign a value to the last (5th) digit. There are now 4 digits remaining to choose from, but one of those values is 0, which we can't choose for the 1st digit. So, we can assign a value to the 1st digit in 3 ways.
At this point, we can assign any of the remaining three digits to the 2nd, 3rd and 4th places. So, we can assign a value to the 2nd digit in 3 ways. We can assign a value to the 3rd digit in 2 ways. We can assign a value to the 4th digit in 1 way. By the Fundamental Counting Principal (FCP), we can complete all five steps (and create a 5 digit number) in (2)(3)(3)(2)(1) ways (= 36 ways)
How many five-digit even numbers can be formed using
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03 Oct 2021, 20:20
Using FCP I. Last digit: 0 Then by FCP, 4x3x2x1=24 II. Last digit: 2 By FCP: 3x3x2x1= 18 III. Last digit: 2 By FCP: 3x3x2x1= 18 SUM OF ABOVE= 24+18+18= 60 Ans: D
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gmatclubot
How many five-digit even numbers can be formed using [#permalink]