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 [imath]2^{10}[/imath] different combinations altogether. Just take away (1+10+45).

 

[Loki123]

There are [imath]2^{10}[/imath] 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.
[math]2^{10}-{10\choose 0}-{10\choose 1}-{10\choose2 }=1024-1-10-45=968[/math]

 

[Loki123]

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

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,
[math](x+y)^n=\sum_{k=0}^{n}{n \choose k}x^ky^{n-k}[/math]Let [imath]x=1,y=1,n=10[/imath] so you have:
[math] (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}} [/math]

 

[Loki123]

From the binomial theorem,
[math](x+y)^n=\sum_{k=0}^{n}{n \choose k}x^ky^{n-k}[/math]Let [imath]x=1,y=1,n=10[/imath] so you have:
[math] (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}} [/math]

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 [imath]2^{10}[/imath].
(Since, for each of the 10 flowers, we can either include it or not, in the arrangement:
giving [imath](\underbrace{2 \times 2 \times 2 \times... \times 2}_{\text{10 times}})[/imath] 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).

[imath]2^{10} -\binom{10}{0} - \binom{10}{1} - \binom{10}{2}[/imath]

 

[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 [imath]\sum\limits_{k = 3}^{10} \dbinom{10}{k}=968 [/imath] 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.