# combination and permutation !!!!

I have 3 questions that i had a try to do but i didn't understand them could anybody please help me to solve these questions.

For Q1 i know how to use the multiplication counting procedures for a) i got 2^7= 128 For b) i got 1x1x1x2x2x2x2=2^4=16 but not sure c) i didn't understand it

Q2 i know that i have to apply the addition principle for counting procedures but i didn't understand the question

Q3 same i didn't understand them but all i know that i have to apply either the combination or permutation but i still cant figure out whats the differences between both of them.

Q1 A binary sequence is a sequence whose elements come from the set {0,1}
a) How many binary sequences of length 7 are there ?
b) How many binary sequences of length 7 begin with 101 ?
c) How many binary sequences of length 7 have at least two 1's ?

Q2 Note that 15876 = 2^2*3^4*7^2

a) Compute the number of distinct divisors of 15876?
b) Compute the number of odd divisors of 15876 ?

Q3 On the menu of a Chinese restaurant there are 6 chicken dishes,
5 beef dishes, 9 seafood dishes and 10 vegetable dishes.

a) In How many ways can a family order if they choose exactly one dish
of each kind ?
b) In how many ways can they order if at most one dish of each kind is chosen ?

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Hello i have removed the questions and updated my Q. – Mora Feb 11 '14 at 21:35

Hint for 1c):

There is exactly $1$ sequence with no $1$ in it and there are $7$ sequences with exacly one $1$ in it. The rest of the sequences all have at least twice a $1$ in it.

2a) you have $15876=2^2\times 3^4\times 7^2$. A divisor has the form $2^a\times 3^b\times 7^c$ with $a\in \{0,1,2\}$ ($3$ choices), $b\in \{0,1,2,3,4\}$ ($5$ choices) and $c\in \{0,1,2\}$ ($3$ choices). So you have $3\times 5\times 3$ choices.

2b) same as in 2a) except that for $a$ you have only one choice: $0$.

I hope that from here someone will take over, because I am going to bed (in Holland it is 22:50) Bye.

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Thank you so much !This was very useful. – Mora Feb 11 '14 at 23:46
You are welcome. – drhab Feb 12 '14 at 8:29

You're right on target for the first two. For the third, start by finding the number of binary sequences of length $7$ with no $1$s , and the number with exactly one $1.$ What can we do then?

A (nonnegative integer) divisor of $15876$ will have $0,1,$ or $2$ factors of $2;$ $0,1,2,3,$ or $4$ factors of $3;$ $0,1,$ or $2$ factors of $7.$ can you take it from there? An odd divisor of $15876$ will have $0$ factors of $2$.

When choosing a dish in the first part of your last question, the family will choose one of $6$ chicken dishes, one of $5$ beef dishes, one of $9$ seafood dishes, and one of $10$ vegetable dishes. Apply fundamental counting principle. As for the second part, the approach is basically the same, but the family may also choose not to order a chicken dish (for example). This gives an additional option in each category. However, they must order something (or at least, the wording of the question suggests this), so we must discard the possibility that they simply stand up and leave without ordering (which our modified count includes as a possibility), so we must subtract $1$ from the answer we seem to get from fundamental counting principle.

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for Q1 C) we will say choose for element for position 1 - 1 element then choices for element for position 2 - 1 . . . . . till position 5,6,7 will have 2 elements is that right ? – Mora Feb 11 '14 at 22:00
For Q3 b) i understand from Q3 A) that every family will choose 1 dish for each but i still cant understand part B which if at most one dish of each kind is chosen ? – Mora Feb 11 '14 at 22:04
I'm not following what you're saying for Q1c, I'm afraid. If a binary sequence has no $1$s, then it has a $0$ in every position. How many such sequences of length $7$ are there? How many binary sequences of length $7$ have a $1$ in exactly one position? I have to sign off for a bit now, so I won't be able to reply to further comments immediately, and will have to address Q3 later. – Cameron Buie Feb 11 '14 at 22:08
Never-mind it make more since now for Q1 c). – Mora Feb 11 '14 at 22:12
Let me put it another way. To choose at most one chicken dish, a family must choose one of the $6$ chicken dishes or none of the chicken dishes. This is $7$ options for the "chicken choice," as opposed to $6$ in part (a). Likewise, there are $6$ options for the "beef choice," $10$ for the "seafood choice," and $11$ for the "vegetable choice." – Cameron Buie Feb 11 '14 at 23:25