Combinations

[Loki123]

There are 10 different flowers. In how many ways can you make a flower arrangement if you have to use at least 3 flowers.
I calculated combinations from 3 flowers to 10. Is there a shorter way?

 

[lex]

There are $2^{10}$ different combinations altogether. Just take away (1+10+45).

 

[Loki123]

There are $2^{10}$ different combinations altogether. Just take away (1+10+45).

That's not correct. 968 is correct

 

[BigBeachBanana]

That's not correct. 968 is correct

Lex's correct, maybe you didn't subtract correctly.
$$2^{10}-{10\choose 0}-{10\choose 1}-{10\choose2 }=1024-1-10-45=968$$

 

[Loki123]

Lex's correct, maybe you didn't subtract correctly.
$$2^{10}-{10\choose 0}-{10\choose 1}-{10\choose2 }=1024-1-10-45=968$$

Oh yeah. But why that way? I don't understand it.

 

[BigBeachBanana]

Oh yeah. But why that way? I don't understand it.

From the binomial theorem,
$$(x+y)^n=\sum_{k=0}^{n}{n \choose k}x^ky^{n-k}$$Let $x=1,y=1,n=10$ so you have:
$$(1+1)^{10}=2^{10}=\sum_{k=0}^{10}{10 \choose k}={10\choose 0}+{10\choose 1}+{10\choose 2}+\dots+{10\choose10}\\ \underbrace{2^{10}-{10\choose 0}-{10\choose 1}-{10\choose 2}}_{\text{Lex's method}}=\underbrace{{10\choose 3}+\dots+{10\choose 10}}_{\text{Your method}}$$

 

[Loki123]

From the binomial theorem,
$$(x+y)^n=\sum_{k=0}^{n}{n \choose k}x^ky^{n-k}$$Let $x=1,y=1,n=10$ so you have:
$$(1+1)^{10}=2^{10}=\sum_{k=0}^{10}{10 \choose k}={10\choose 0}+{10\choose 1}+{10\choose 2}+\dots+{10\choose10}\\ \underbrace{2^{10}-{10\choose 0}-{10\choose 1}-{10\choose 2}}_{\text{Lex's method}}=\underbrace{{10\choose 3}+\dots+{10\choose 10}}_{\text{Your method}}$$

But why 1+1 if we need to use at least 3 flowers

 

[lex]

The total number of combinations using any number of flowers is $2^{10}$.
(Since, for each of the 10 flowers, we can either include it or not, in the arrangement:
giving $(\underbrace{2 \times 2 \times 2 \times... \times 2}_{\text{10 times}})$ possible outcomes.

As you say, we need to have at least 3 flowers, so we eliminate the unwanted combinations (with 0 flowers, 1 flower, or 2 flowers).

$2^{10} -\binom{10}{0} - \binom{10}{1} - \binom{10}{2}$

 

[pka]

There are 10 different flowers. In how many ways can you make a flower arrangement if you have to use at least 3 flowers.
I calculated combinations from 3 flowers to 10. Is there a shorter way? View attachment 32242

Your work is correct $\sum\limits_{k = 3}^{10} \dbinom{10}{k}=968$ SEE HERE
Good work!

 

[Steven G]

But why 1+1 if we need to use at least 3 flowers

We let x=y=1 so (x+y)ⁿ=$\displaystyle (x+y)^n=\sum_{k=0}^{n}{n \choose k}x^ky^{n-k}$ becomes $\displaystyle 2^n=\sum_{k=0}^{n}{n \choose k}$

The 1+1 has nothing to do with the number of flowers you are picking. It is just to get rid of $\displaystyle x^ky^{n-k}$

 

[Loki123]

We let x=y=1 so (x+y)ⁿ=$\displaystyle (x+y)^n=\sum_{k=0}^{n}{n \choose k}x^ky^{n-k}$ becomes $\displaystyle 2^n=\sum_{k=0}^{n}{n \choose k}$

The 1+1 has nothing to do with the number of flowers you are picking. It is just to get rid of $\displaystyle x^ky^{n-k}$

So then aren't we just doing combinations without repetition?

 

[BigBeachBanana]

So then aren't we just doing combinations without repetition?

Why do you think it's with repetition? You have 10 different flowers.

 

[Loki123]

Why do you think it's with repetition? You have 10 different flowers.

i said without.

 

[BigBeachBanana]

i said without.

You answered your own question.