How many five-digit numbers can be made from the digits 1 to 7 if repetition is allowed

Your formula is the number of distinct numbers that can be created from all of those digits. If we only want five, we’re going to have to do casework.

The set of digits for this five digit number must look either like $\{w,w,x,y,z\}$, like $\{w,w,x,x,y\}$, or like $\{w,w,w,x,x\}$. For each of these cases, we can first count the number of ways to choose the digits, and then multiply by the number of ways to order them. This gives:

• $\{w,w,x,y,z\}$: $3\cdot\frac{5!}{2!\cdot1!\cdot1!\cdot1!}=180$.
• $\{w,w,x,x,y\}$: $6\cdot\frac{5!}{2!\cdot2!\cdot1!}=180$.
• $\{w,w,w,x,x\}$: $2\cdot\frac{5!}{3!\cdot2!}=20$.

This gives a total of $\boxed{380}$ numbers.

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Solution

The correct option is B 1800As two is already included, four more digits are required to be selected out of 6 digits. The number of ways of selecting the numbers = 6C4 = 6C2 = 15 ways Also, five digits can be arranged amongst themselves in 5! = 120 ways. Hence the total no. of possibilities =120×15=1800 ways.

How many five-digit numbers can be made from the digits 1 to 7 if repetition is allowed?

(a) 16807

(b) 54629

(c) 23467

(d) 32354

This question was addressed to me during an internship interview.

My question is from Fundamental Principle of Counting in portion Counting of Discrete Mathematics

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