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Bunuel
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Olive is creating a five-digit code using the digits 0 through 9. How 13 May 2021, 04:15
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D
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65% (02:28) correct 35% (02:57) wrong based on 264 sessions

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Olive is creating a five-digit code using the digits 0 through 9. How many different codes can she create with exactly two prime digits if no digits can be repeated?

A. 252

B. 3,120

C. 3,456

D. 14,400

E. 30,240
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D
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Olive is creating a five-digit code using the digits 0 through 9. How 16 May 2021, 08:00
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Bunuel
Olive is creating a five-digit code using the digits 0 through 9. How many different codes can she create with exactly two prime digits if no digits can be repeated?

A. 252

B. 3,120

C. 3,456

D. 14,400

E. 30,240
Show more

The code will be any combination of the following: {P1} {P2} {X} {Y} {Z}, where P1 and P2 are prime digits, and X, Y and Z are other digits (all distinct).

There are four single-digit primes: 2, 3, 5, and 7 and six other digits: 0, 1, 4, 6, 8, and 9.

The number of pairs of primes, P1 and P2, out of four primes is 4C2 = 6;
The number of triplets of non-primes, X, Y and Z, our of six is 6C3 = 20.

Therefore, 6*20*5! = 14,400 (multiplying by 5! to account for different arrangements of 5 digits {P1} {P2} {X} {Y} {Z}).

Answer: D.
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Olive is creating a five-digit code using the digits 0 through 9. How 13 May 2021, 04:32
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Bunuel
Olive is creating a five-digit code using the digits 0 through 9. How many different codes can she create with exactly two prime digits if no digits can be repeated?

A. 252

B. 3,120

C. 3,456

D. 14,400

E. 30,240
Show more


The five digit code consists of.
1) 2 prime digits:
There are 4 prime digits: 2,3,5,7. So 4C2=6
2) 3 from remaining digits:
There are 10-4 or 6 remaining out of which we have to choose 3. So 6C3=20

Now these five digits can be arranged in 5! or 120 ways. As this is a code, 0 can take any position. 05874 will be correct.

Total 6*20*120=14400

D
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Olive is creating a five-digit code using the digits 0 through 9. How 16 May 2021, 09:04
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Olive is creating a five-digit code using the digits 0 through 9. How many different codes can she create with exactly two prime digits if no digits can be repeated?

Selection of Prime nos. = 4C2 =6 ways
Selection of other digits = 6C3 = 20 ways
And arranged in 5! =120 ways

So, Total nos. of ways = 6*20*120= 14400

I think D. :)
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Olive is creating a five-digit code using the digits 0 through 9. How 17 May 2021, 00:14
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Total numbers: 10 [0 to 9]
Prime numbers: 4 [2, 3, 5, 7]
Code: 5 digits: _, _, _, _, _
Condition: Repetition not allowed and exactly 2 prime needed.

First select two primes out of 4: \(^4{C_2} = 6\). These 6 ways will cover the place for 2 digits.

Remaining 6 digits will be selected from 6 digits: \(^6{C_3} = 20\).

To arrange these 5 digits, we need \(5! = 120\) ways.

Total ways: \(6 * 20 * 120 = 14,400\)

Answer D
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Olive is creating a five-digit code using the digits 0 through 9. How 27 Apr 2025, 02:28
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1. Thanks for the explanation. I'm still not able to understand why at last you have multiplied by 5!

2. Also, I solved this question in the following way and got wrong. Can you pls explain where i lost my track or is my approach fundamentally wrong.

There are 5 digits, So _ _ _ _ _
There are 4 prime numbers between 0 to 9.
There are 6 non-prime numbers.
Having 2 prime numbers is a must.

Therefore by using fundamental counting principle and filling the above blanks.

4 3 6 5 4 (The first two numbers are filled with prime, remaining three are non prime)

Therefore, 4x3x6x5x4 = 1440.

Now, prime numbers can be arranged in the following.

P1 NP P2 NP NP - Keeping the first number in position, the second number can be swapped 4 times. Now, move the first number (P1) to second slot and keep rearranging second number (P2). So, you get 4x5 = 20 possibilities.

So, I thought 1440 x 20 = 28800.

P.S. Pls excuse if I got the logic terribly wrong.

Bunuel
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Olive is creating a five-digit code using the digits 0 through 9. How many different codes can she create with exactly two prime digits if no digits can be repeated?

A. 252

B. 3,120

C. 3,456

D. 14,400

E. 30,240

The code will be any combination of the following: {P1} {P2} {X} {Y} {Z}, where P1 and P2 are prime digits, and X, Y and Z are other digits (all distinct).

There are four single-digit primes: 2, 3, 5, and 7 and six other digits: 0, 1, 4, 6, 8, and 9.

The number of pairs of primes, P1 and P2, out of four primes is 4C2 = 6;
The number of triplets of non-primes, X, Y and Z, our of six is 6C3 = 20.

Therefore, 6*20*5! = 14,400 (multiplying by 5! to account for different arrangements of 5 digits {P1} {P2} {X} {Y} {Z}).

Answer: D.
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Olive is creating a five-digit code using the digits 0 through 9. How 27 Apr 2025, 03:06
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1. Thanks for the explanation. I'm still not able to understand why at last you have multiplied by 5!

2. Also, I solved this question in the following way and got wrong. Can you pls explain where i lost my track or is my approach fundamentally wrong.

There are 5 digits, So _ _ _ _ _
There are 4 prime numbers between 0 to 9.
There are 6 non-prime numbers.
Having 2 prime numbers is a must.

Therefore by using fundamental counting principle and filling the above blanks.

4 3 6 5 4 (The first two numbers are filled with prime, remaining three are non prime)

Therefore, 4x3x6x5x4 = 1440.

Now, prime numbers can be arranged in the following.

P1 NP P2 NP NP - Keeping the first number in position, the second number can be swapped 4 times. Now, move the first number (P1) to second slot and keep rearranging second number (P2). So, you get 4x5 = 20 possibilities.

So, I thought 1440 x 20 = 28800.

P.S. Pls excuse if I got the logic terribly wrong.

Bunuel
Bunuel
Olive is creating a five-digit code using the digits 0 through 9. How many different codes can she create with exactly two prime digits if no digits can be repeated?

A. 252

B. 3,120

C. 3,456

D. 14,400

E. 30,240

The code will be any combination of the following: {P1} {P2} {X} {Y} {Z}, where P1 and P2 are prime digits, and X, Y and Z are other digits (all distinct).

There are four single-digit primes: 2, 3, 5, and 7 and six other digits: 0, 1, 4, 6, 8, and 9.

The number of pairs of primes, P1 and P2, out of four primes is 4C2 = 6;
The number of triplets of non-primes, X, Y and Z, our of six is 6C3 = 20.

Therefore, 6*20*5! = 14,400 (multiplying by 5! to account for different arrangements of 5 digits {P1} {P2} {X} {Y} {Z}).

Answer: D.
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The problem is when you multiply by 20. After doing (4*3)(6*5*4), 4*3 already gives ordered primes (so 2! is already accounted for) and 6*5*4 already gives ordered non-primes (so 3! is already accounted for). Now to arrange P, P, not P, not P, not P, you need to multiply by 5!/(2!*3!) = 10, not by 20. That is where you went wrong.
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Olive is creating a five-digit code using the digits 0 through 9. How 29 Apr 2025, 04:34
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Wouldn't it be a permutations problem? Since the order of the numbers matter?
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Olive is creating a five-digit code using the digits 0 through 9. How 29 Apr 2025, 06:13
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Bunuel
Bunuel
Olive is creating a five-digit code using the digits 0 through 9. How many different codes can she create with exactly two prime digits if no digits can be repeated?

A. 252

B. 3,120

C. 3,456

D. 14,400

E. 30,240

The code will be any combination of the following: {P1} {P2} {X} {Y} {Z}, where P1 and P2 are prime digits, and X, Y and Z are other digits (all distinct).

There are four single-digit primes: 2, 3, 5, and 7 and six other digits: 0, 1, 4, 6, 8, and 9.

The number of pairs of primes, P1 and P2, out of four primes is 4C2 = 6;
The number of triplets of non-primes, X, Y and Z, our of six is 6C3 = 20.

Therefore, 6*20*5! = 14,400 (multiplying by 5! to account for different arrangements of 5 digits {P1} {P2} {X} {Y} {Z}).

Answer: D.
Wouldn't it be a permutations problem? Since the order of the numbers matter?
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Yes, it is a permutation problem. That is why after choosing the digits (2 primes and 3 non-primes), we multiply by 5! to account for all possible orders. The combinations (4C2 and 6C3) handle the selection of digits, and 5! handles the different possible arrangements.
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