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Problem 2: Partitions of 10 and 20Show that You can make up 10 pence in eleven ways using IOp, Sp, Zp and Ip coins:(ii) In how many ways can you make Up Z0 pence us...

Question

Problem 2: Partitions of 10 and 20Show that You can make up 10 pence in eleven ways using IOp, Sp, Zp and Ip coins:(ii) In how many ways can you make Up Z0 pence using Z0p, '10p, Sp, 2p and Ip coins?

Problem 2: Partitions of 10 and 20 Show that You can make up 10 pence in eleven ways using IOp, Sp, Zp and Ip coins: (ii) In how many ways can you make Up Z0 pence using Z0p, '10p, Sp, 2p and Ip coins?



Answers

These problems involve distinguishable permutations. Arranging Coins In how many different ways can four pennies, three nickels, two dimes, and three quarters be arranged in a row?

Okay, so we have, um, five different kinds of coins and we have to pay place these 20 coins into the different configuration is So, for example, in the sequence, I have four pennies. Three dimes. No, 35 cents coins without one extra cute to then sends going on. That was me. Four quarters and seven. How? Dollar coins. So what we want is to know how to place these 20 coins in essence sequence off 24 elements. So we have 24 spots, and we're here to place 20 coins when I know how to do that, That's Tony for community aerial number with 20. And that's the result. If we calculate that not.

To solve this it prison for X, we can write X ical do minus B plus minus or neutral the square minus four a. C upon the way. So export the values off a Lee and see from the Warwick is and really get X equal to minus off. Mine ist away plus minus our little for way square minus four times is for And then see civilly three way square minus two provolone, two m's or which is to it so further. Really? Here ex acquittal The way Bless minus are literally four way square minus or deer way squared plus 32 Oh, on it. Which you may go away. Bill s minus on literal 32 minus form before. Why Square of own. Okay, so this will be the value for X. It's leaking right next door, the way plus minus in the root 30 to minus 44 way square a phone it. And that's all that prison for way. We can fight it as de y square minus two x ray Plus for excess square minus CO equal to zero. So Villiger Oh, I could do minus be less minus a little The square menace racy upon the way which will be I am. Missile minus two X plus minus are literal. Four X is grand minus core EMS. Every just three earned Sees for excess square minus two. Phone. We then really get who wicks plus minus on the roof or excess square minus or ph. X is glad plus 24 of own six. Then we will hear why you could do do X less minus. Earn General 24 minus 44 Xs Square on six so well to going from this record then and then. Further Hilger X bless minus a little six minus level nexus square upon three. So the well you for Why will be X plus minus a little six minus 11 excess square of our three? So this will be the value off. Why in terms off X

Hey, it's clear. So when you read here, So let's make X up. One the number of pennies selected and except two me the number of nickels selected. So therefore six of one must accept to is equal to eat, and they both have to be equivalent or greater than one. So the number of selections is gonna be equal to the number of, um, entered your solutions. So this is equivalent thio plus to minus one choose eight, which is equivalent to nine choose eight, which is equal to nine.

You have two nickels, one dime and one half dollar. And how many ways can we select one or more coins? Let's look at some characteristics of this problem. One characteristic is that as we select the coins, order is not important. Another characteristic is then two off the coins are identical. Let's look at some formulas that we can potentially use. So let's look at this formula combination and taken our. The assumption here is that we have in different objects. The juice from and order is not important, but in our problem, two of the objects took. The coins are identical, so we don't have four different objects, so we cannot use this formula about permutation. Rule number two. This formula here would you have n objects? Where are one are identical, arto are identical and so on. Now it's a permutation formula, which means order. It's important, so we cannot use that. In fact, the best way to solve this problem is just by listing all the different ways and then just counting the number of ways. So let's try doing that. So when we say one or more coins are to be selected and that means one coin for two coins or three coins or four points are selected. Let's let down the possibilities for each. Okay, now the ways to select one coin would be one nickel one time. Or you could select one half dollar. Note that the two nickels are identical, so they're indistinguishable. Ways to select two coins would be two nickels, one nickel, one dime, one nickel one have dollar, one dying, one half dollar ways to select three coins would be do a nickel one time do nickel one half dollar, one nickel, one dime, one half dollar. There's only one way to select all four points. Do Nico one dime, one half dollar. So let's can't the number off wastes to select the coins one or more going suits. 1234567 89 10 11. So there are 11 ways to select one or more coins from two nickels, one dime and one half brother


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