How many five-digit numbers are there, if the two leftmost

Question

How many five-digit numbers are there, if the two leftmost digits are even, the other digits are odd and the digit 4 cannot appear more than once in the number.

A. 1875
B. 2000
C. 2375
D. 2500
E. 3875

walker · Accepted Answer

C.

N=(4*5-1)*5*5*5=2375

where
4 cases of first digit {2,4,6,8} 
5 cases of second digit {0,2,4,6,8} 
1 case of 44 for two leftmost digit
5 cases of third digit {1,3,5,7,9}
5 cases of fourth digit {1,3,5,7,9}
5 cases of fifth digit {1,3,5,7,9}

x2suresh · Answer

EEOOO

= (combinations excludes 4 in first poistion) + (combinations with first position 4)
 = (3*5)*(5*5*5)+(1*4)*5*5*5 
 = 19*(5^3)= 2375

Bunuel · Answer

How many five-digit numbers are there, if the two leftmost digits are even, the other digits are odd and the digit 4 cannot appear more than once in the number.

A. 1875
B. 2000
C. 2375
D. 2500
E. 3875

Easier approach:

Total five-digit numbers with two even leftmost digits and with three odd last digits: EEOOO - \(4*5*5*5*5=4*5^4\) (notice that we have only 4 choices for the first digit since we cannot use zero);

Total five-digit numbers with two 4's as leftmost digits and with three odd last digits: 44OOO - \(5*5*5=5^3\);

So, the answer is \(4*5^4-5^3=5^3(4*5-1)=125*19=2375\).

Answer: C.

Similar question to practice:
https://gmatclub.com/forum/how-many-5-di ... 26966.html
https://gmatclub.com/forum/how-many-5-di ... 88-20.html

Hope it helps.

GK_Gmat · Answer

Very nice. I was stupid enough to make the mistake of forgetting 44 (4*4*5*5*5 = 2000). Walker you're an ace in quant buddy - how the hell do you study for it? It'll be great if you start a topic 'My Quant Prep Strategies' on the 'GMAT Strategies' section

walker · Answer

Thanks, GK_Gmat. you are all here very nice in GMAT :-D

. Unfortunately, I guess that my quant preparation couldn't help you.

. 95% of my preparation - studying in past. I finished physicals mathematical school, have Master of Science in Physics, Master of Science in Finance, and PhD in nannotech.... But I'll try to help anyone on this forum and I hope you will help me in Verbal :wink:

MBAhereIcome · Answer

there are three possible patters.

1. when 4 is the 1st digit, and not the 2nd digit  = 1*4*5*5*5 = 4*5^3 
2. when 4 is the 2nd digit, and not the 1st digit  = 3*1*5*5*5 = 3*5^3  ... 1st digit can't be 0 either
1. when neither 1st nor 2nd digit is 4 = 3*4*5*5*5 = 12*5^3

total =  19*5^3 = 2375

SVaidyaraman · Answer

First constraint=2 leftmost digits even. Therefore 4*5 ways of forming the 2 leftmost digits as the number cannot start with 0.
Constraint within this constraint: 4 should not occur twice. We see that 4 can occur twice only once . Therefore tre are 4*5 - 1 ways of forming the 2 leftmost digits=19 ways
Third constraint: The other 3 digits are odd. Therefore 5*5*5 ways of forming the 3 rightmost digits=125 ways

Total number of possible combinations given the constraints is 19*125= 2375

Fdambro294 · Answer

Constraints:

must be a 5 Digit Numbers

10,000s digit must be EVEN

1,000s digit must be EVEN

Last 3 Digits must be ODD

and

4 can only be used ONE TIME as a Digit in any given number

(Step 1) Break the possible numbers into completely exhaustive cases that cover every possibility

For every possible number, it must fall under 1 of 3 characterizations:

Case 1: 4 is the Digit in the 10,000s Place

OR

Case 2: 4 is the Digit in the 1,000s Place

OR

Case 3: 4 is NOT used in EITHER of the two above Places

Case 1:

10,000s Place: Fix the Digit 4 in this place for every possible arrangement of Digits ----- 1 option

1,000s Place: Must be an EVEN Digit, but we can not repeat 4 ------ 0 or 2 or 6 or 8 ------ 4 Options

100s Place:

10s Place:

Units Place:

Since Repeatition of the digits is allowed and since the last 3 digits have to include an ODD Digit, we have 5 Available options for each Place ------ 1 or 3 or 5 or 7 or 9

Case 1 Count of Arrangements:

(1) * (4) * (5)^3

Case 2:

10,000s Place: since we are fixing 4 in the next Place and the Numbers must be 5 Digit Numbers, we can NOT use the Even Digits of 0 and 4 -------- 2 or 6 or 8 ----- 3 Available Options

1,000s Place: Fix the Digit 4 in this place ------ 1 Options

Last 3 Places: Each Place has 5 available options ---- the Odd Digits ---- 5 options

Case 2 Count of Arrangements: (3) * (1) * (5)^3

Case 3: neither the 10,000s Place nor the 1,000s Place has a Digit 4

10,000s Place: can NOT have the Even Digits 4 and 0 ------ 2 or 6 or 8 ----- 3 available options

1,000s Place: can NOT have the Even Digit 4 for this scenario ---- 0 or 2 or 6 or 8 ---- 4 available options

Last 3 Digits: Each place can again have any one of 5 ODD Digits ---- 5 options for each space

(3) * (4) * (5)^3 = (12) * (5)^3

Sum up the 3 Cases:

(4) * (5)^3 + (3) * (5)^3 + (12) * (5)^3 =

(5)^3 *   =

(5)^3 * 19 = (125 * 20) - (125 * 1) = 2,500 - 125 =

2,375 (C)

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How many five-digit numbers are there, if the two leftmost

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