Compute the number of ways to arrange 7 of those cookies into a straight line.

Question

You baked 6 indistinguishable snickerdoodle cookies and 8 indistinguishable chocolate chip cookies. Compute the number of ways to arrange 7 of those cookies into a straight line.

My try: First, identify the number of snickerdoodle cookies, denoted as 𝑠, and the number of chocolate chip cookies, denoted as c, such that $s+c=7$. The values of s can range from $0$ to $6$ (since there are only 6 snickerdoodle cookies available).

For each possible value of s, the corresponding value of c is given by $c=7−s$. Also, c must not exceed 8 (the number of available chocolate chip cookies).

Next, we compute the number of distinct arrangements for each valid combination of s and c using the binomial coefficient, which counts the number of ways to choose s positions out of 7 for the snickerdoodle cookies. This is given by: $C(7,s)$ ​

Let's calculate it for each valid s we get, $$\sum_{s=0}^7C(7,s)=1+7+21+35+35+21+7=127$$ Thus, the total number of ways to arrange 7 cookies from 6 Snickerdoodle cookies and 8 chocolate chip cookies is $127$.

I know that the approach is not correct and I need to use the indistinguishable one differently. Can you help me with a well-explained solution? ​ I was also trying to think about putting a divider between cookies but couldn't get a good idea. Any help!

Answer 1

I think the solution is correct. I got the same answer, and here's how I did it:

We're looking for the number of sequences of 7 cookies, where 2 types of cookies exist. There's {2^7=128} ways to do this, but we exclude the 7 snickerdoodle possibility, so we have $\boxed{127}$.

Could you check if you misinterpreted the question, or perhaps verify again whether the answer is wrong?

Answer 2

Here's another approach that seems to work for this case.

First, suppose we had an infinite supply of both types of cookie.  In that case, we'd have a free choice of either type of cookie for the first one, either type for the second, and so on until the 7th.  So that would give 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2⁷ = 128 ways.

However, that counts some ways that aren't allowed with our finite supply of cookies.  The chocolate cookies aren't a limitation, because we have more than enough to have all 7 chocolate.  But we can't have all 7 snickerdoodle, because there are only 6.

That disallows one single way — all 7 snickerdoodle — and so the final answer is 2⁷ - 1 = 128 - 1 = 127 ways.

(That's an easier calculation, though it's probably less general, and would have more trouble if there were fewer cookies available.)

Answer 3

I am new to combinatorics and I guess this might be an answer, not sure though: To find the number of ways to arrange $7$ cookies out of $6$ snickerdoodle cookies and $8$ chocolate chip cookies, we can use the binomial coefficient.

Given:

• $6$ snickerdoodle cookies (S)
• $8$ chocolate chip cookies (C)
• We need to choose $7$ cookies out of these 14 in total.

We can represent the problem as selecting a combination of snickerdoodle and chocolate chip cookies that add up to $7$.

Let $k$ be the number of snickerdoodle cookies chosen, and $7-k$ be the number of chocolate chip cookies chosen.

The possible values for $k$ range from $0$ to $6$ (since there are only $6$ snickerdoodle cookies available).

For each $k$, we compute the number of ways to choose $k$ snickerdoodle cookies from $6$, and $7-k$ chocolate chip cookies from $8$.

The number of ways to choose $k$ snickerdoodle cookies from 6 is $\binom{6}{k} $.

The number of ways to choose $ 7 - k $ chocolate chip cookies from 8 is $ \binom{8}{7 - k} $.

The total number of ways for each $ k $ is the product of these two binomial coefficients.

So, the total number of ways to arrange 7 cookies is the sum of these products for all possible values of $ k $:

$$\sum_{k=0}^{6}\binom{6}{k}\binom{8}{7-k}$$

If you compute this sum, the number of ways to arrange $7$ cookies from a selection of $6$ snickerdoodle cookies and $8$ chocolate chip cookies is $ 3432 $.