# Thread: 5 Basic tastes, how many combinations?

1. ## 5 Basic tastes, how many combinations?

Question: If the papillae (bumps on your tongue) can sense 5 basic tastes and any combination of those tastes produce a unique taste, how many possible unique taste can you have?

Misconception: 4+3+2+1 = 10 unique tastes, but that only accounts for non-repetitive pairs of tastes. In this problem, any number of the 5 basic, different taste can account for 1 unique taste (e.g., taste A,B,C can mix to create 1 unique taste). With that in mind, how many possible unique taste can you have?

My 2nd answer: 16 unique tastes.
(4+3) + (3+2) + (2+1) + 1

2. Originally Posted by Masterthief1324
Question: If the papillae (bumps on your tongue) can sense 5 basic tastes and any combination of those tastes produce a unique taste, how many possible unique taste can you have?

Misconception: 4+3+2+1 = 10 unique tastes, but that only accounts for non-repetitive pairs of tastes. In this problem, any number of the 5 basic, different taste can account for 1 unique taste (e.g., taste A,B,C can mix to create 1 unique taste). With that in mind, how many possible unique taste can you have?

My 2nd answer: 16 unique tastes.
(4+3) + (3+2) + (2+1) + 1

I get 31 unique tastes.

3. Well It should be $\displaystyle 31$

$\displaystyle 31 = {5 \choose 1}+ {5\choose 2}+{5\choose 3}+{5\choose 4}+{5\choose 5}$

(Is "no taste" a taste too? ;p)

4. Originally Posted by Dinkydoe
(Is "no taste" a taste too? ;p)

5. Hello, Masterthief1324!

I agree with skeeter and Dinkydoe . . .

If the papillae (bumps on your tongue) can sense 5 basic tastes
and any combination of those tastes produce a unique taste,
how many possible unique tastes can you have?

. . $\displaystyle \begin{array}{ccccc} \text{1-at-a-time:} & _5C_1 &=& 5 \\ \text{2-at-a-time:} & _5C_2 &=& 10 \\ \text{3-at-a-time:} & _5C_3 &=& 10 \\ \text{4-at-a-time:} & _5C_4 &=& 5 \\ \text{5-at-a-time:} & _5C_5 &=& 1 \\ \hline & \text{Total:} && {\color{blue}31} \end{array}$

Call the five basic tastes: $\displaystyle a,b,c,d,e$
You can list them and count them yourself . . .

. . $\displaystyle \begin{array}{ccc}\text{1-at-a-time:} & a,b,c,d,e \\ \\ \text{2-at-a-time:} & ab,ac,ad,ae,bc\\ & bd,be,cd, ce, de \\ \\ \text{3-at-a-time:} & abc, abd, abe, acd, ace \\ & ade, bcd, bce, bde, cde \\ \\ \text{4-at-a-time:} & abcd, abce, abde, acde, bcde \\ \\ \text{5-at-a-time:} & abcde \end{array}$

6. Originally Posted by Soroban
Hello, Masterthief1324!

I agree with skeeter and Dinkydoe . . .

. . $\displaystyle \begin{array}{ccccc}$$\displaystyle \text{1-at-a-time:} & _5C_1 &=& 5 \\ \text{2-at-a-time:} & _5C_2 &=& 10 \\ \text{3-at-a-time:} & _5C_3 &=& 10 \\ \text{4-at-a-time:} & _5C_4 &=& 5 \\ \text{5-at-a-time:} & _5C_5 &=& 1 \\ \hline & \text{Total:} && {\color{blue}31} \end{array} Call the five basic tastes: \displaystyle a,b,c,d,e You can list them and count them yourself . . . . . \displaystyle \begin{array}{ccc}\text{1-at-a-time:} & a,b,c,d,e \\ \\$$\displaystyle \text{2-at-a-time:} & ab,ac,ad,ae,bc\\ & bd,be,cd, ce, de \\ \\ \text{3-at-a-time:} & abc, abd, abe, acd, ace \\ & ade, bcd, bce, bde, cde \\ \\ \text{4-at-a-time:} & abcd, abce, abde, acde, bcde \\ \\ \text{5-at-a-time:} & abcde \end{array}$

I will familiarize myself with the combination and permutation notation later.
If I was to write down all the possible combinations, what is the best way to keep track of the combinations so as to avoid overlapping combinations?

7. Originally Posted by Masterthief1324
I will familiarize myself with the combination and permutation notation later.
If I was to write down all the possible combinations, what is the best way to keep track of the combinations so as to avoid overlapping combinations?
No "best" way ... just be organized.

Soroban's listing is very organized.

8. Here is another way to arrive at the same answer.

Each of the 5 tastes is either present or not present-- 2 choices. So there are

$\displaystyle 2^5$

possibilities. But one of those is the "no taste" combination. If we eliminate that one, there are

$\displaystyle 2^5 -1 = 31$

possibilities.