MMLU-Pro Question 7759

Question

In how many ways can we form a 7-digit number using the digits 1, 2, 2, 3, 3, 3, 3?
A) 5040
B) 2520
C) 105
D) 720
E) 420
F) 120
G) 840
H) 315
I) 210
J) 60
Correct Answer: C
0% correct
100% correct

Model Reasoning

 Okay, so I need to figure out how many different 7-digit numbers we can make using the digits 1, 2, 2, 3, 3, 3, 3. Let me think. Hmm, this seems like a permutation problem with repeated elements. First, I remember that when you have a set of items where some are identical, the number of distinct permutations is calculated by dividing the factorial of the total number of items by the product of the factorials of the counts of each repeated item. The formula is n! divided by (k1! * k2! * ... * km!), where n is the total number of items, and k1, k2, etc., are the counts of each repeated element. In this case, the digits we have are: one 1, two 2s, and four 3s. Let me check: 1 + 2 + 4 equals 7 digits total, which matches the 7-digit number we need. So the total number of permutations should be 7! divided by (1! * 2! * 4!). Let me compute that. First, 7! is 5040. Then, 1! is 1, 2! is 2, and 4! is 24. So the denominator is 1 * 2 * 24 = 48. Then, 5040 divided by 48. Let me do that division. 5040 divided by 48. Let's see: 48 times 100 is 4800. 5040 minus 4800 is 240. 240 divided by 48 is 5. So total is 100 + 5 = 105. Wait, so 105? Let me check again. 48 * 105 = 5040. Yes, 48*100=4800, 48*5=240, so 4800+240=5040. Correct. So the answer should be 105. Looking at the options, option C is 105. So the answer is C.