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Simple counting exercise, with combinations

(a) # of ways to choose string $=$ $\\displaystyle \\prod\\limits_{i=1}^{\\var{n}}$ (# ways to choose $i$th digit) $=$ $\\displaystyle \\prod\\limits_{i=1}^{\\var{n}} 9 = {9^\\var{n} }$.

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(b) # of ways to choose string $=$ (# of ways of choosing an even digit) $\\times$ (# of ways of choosing an odd digit) $\\times \\displaystyle \\prod\\limits_{i=3}^{\\var{n}}$ (# ways to choose a digit less than $\\var{b}$) $=$ $4 \\times 5 \\times {\\var{b-1}^\\var{n-2}}$.

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(c) # of ways to choose string $=$ (# of ways of placing the 9s) $\\times$ (# of ways of placing the remaining digits) $=$ ${C(\\var{n},\\var{c})}\\times 8^\\var{n-c}$.

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A randomised string of $\\var{n}$ digits is made using only the non-zero digits $1,2,3,4,5,6,7,8,9$.

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When each digit is chosen independently, the number of ways to choose the string is just the product of the number of ways to choose each of the digits.

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# of ways to choose string = $\\displaystyle \\prod\\limits_{i=1}^{\\var{n}}$ (# ways to choose $i$th digit).

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It might be helpful to first consider some examples. The chosen string could be

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$\\var{ex1}$ or $\\var{ex2}$.

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How many ways can you choose the first digit?

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How many ways can you choose the $\\var{n}$th digit?

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Assuming there are no other restrictions, how many strings are possible?

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How many ways can an even number be chosen for the first digit?

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How many ways can an odd number be chosen for the second digit?

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How many ways can the remaining $\\var{n-2}$ digits be chosen, given that they must each be less than $\\var{b}$?

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How many strings are possible if the first digit must be even, the second digit odd, and the remaining $\\var{n-2}$ digits are less than $\\var{b}$?

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Note that each digit is no longer independently chosen. But we can split the question into two independent parts.

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First, count the number of ways to choose positions for each of the $\\var{c}$ nines. Since the order of the positions does not matter (all the nines look identical), this is a combination. Remember you can use the NUMBAS command comb(n,r) for $C(n,r)$.

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Then, count the number of ways to choose the digits for the rest of the string. Remember, you can't use the digit nine anymore, so there are 8 choices for each of the remaining unfilled positions.

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How many strings are possible which contain exactly $\\var{c}$ nines?

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