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I have $7$ yoghurts to last the week. Two are strawberry, three raspberry and two peach. I select one yoghurt each day. In how many different orders can I eat the yoghurts?

If I select a yoghurt at random each day, what is the probability that I eat the two strawberry ones on consecutive days?

I thought just going for $7!$ would work but I got the wrong answer. How do I figure this out?

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    You must divide by the ways to reorder the indistinguishable ones. This happens because the $7!$ doesn't account for the fact that if you swap the two peach ones, you get the same order. – 2017-01-12
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    Also, please type your question instead of inserting a picture of it. – 2017-01-12
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    The problem doesn't explicitly say that any two of the same flavor are indistinguishable, but it seems to make sense only if that is what was meant. – 2017-01-12
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    The answer to the first question is $\frac{(2+3+2)!}{2!\times3!\times2!}$. The answer to the second question is $\frac{6!\times2!}{7!}$. – 2017-01-13

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The correct result is $$\frac{7!}{3!\cdot2!\cdot2!}=210$$

You have to consider that changing elements with the same name does not change the combination. We can arrange the three raspberry, the two strawberry and the two peach yoghurts in $3!\cdot 2!\cdot 2!$ ways that lead to identical combinations. Therefore we must divide by this number.

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As explained by Peter, for the first part of the question, the answer is $\dfrac{7!}{3!2!2!} = 210$

For the second part of the question,
two strawberry types can be together in $6$ ways (days $1-2\;thru\;6-7$)
The rest can be consumed in $\dfrac{5!}{3!2!} = 10$ ways,
thus overall number of ways will be $6\times10 = 60$ ways

and $Pr = \dfrac{60}{210} = \dfrac27$