Need help with some permutation and combination problems.?

Question

1.  How many ways can I give 10 identical candies to 4 children?  It s okay if some children don t receive any at all.

2.  I m superstitious.  I want to know how many odd, four-digit numbers there are, but I don t want to count any with a "13" in them (1 and 3 right next to each other)?  How many are there?

3.  I roll two six-sided dice, one white and one black.  How many different rolls could result where the black die shows a greater value than the white die?  How many possible rolls where the two dice together total 7?

4.  An organization desires to select a President and Vice President for an Executive Committee, but both must be from the same division.  There are 17 candidates from the Eastern Division, and 24 from the Western Division.  How many ways could we choose a Pres and VP?

5.  How many permutations of the letters in the word COMPUTER end in a vowel?

6.  How many ways are there to seat 8 men and 4 women in a row if all the women must sit together?

7.  I d like to host a dinner party.  I have 14 friends, but I can only invite 6.  Two of my friends hate each other, and it would be a disaster to invite them both.  How many ways could I make my guest list?

8.  A snail starts in the northwest square of a chess board (8 x 8) and moves east or south at random, one square at a time, until reaching the southeast square.  How many possible paths could the snail take?

Nick · Answer

If I give you the calculations, can you do the rest?

1. C(13,3)

Count arrangements of 10 candies and 3 dividers separating the 4 children's candies 
c c c c c c c c c c | | |

2. 5*9*10^2 - 5*10 - 9*10 - 5*9 + 1

5 odd digits for units, 9 digits for 1000's and 10 digits for 100s and 10s. Subtract 5*10 cases where 13_ _, 9*10 cases where _ _ 13 and 5*9 cases where _ 13 _, add back the single overlap case 1313.

3. (6^2 - 6)/2

There are 6^2 total outcomes (6 for each die), subtract the 6 cases where both dice show the same result then /2 since black die is greater equally as often as white die.

4. P(17,2) + P(24,2)

Permutations of P and VP (2 positions) from 17 people, *or* permutations of P and VP from 24. Add these two cases.

5. 3*7!

3 vowels can go in last place, for each there are 7 distinct letters to arrange in the first 7 places in 7! ways.

6. 9!4!

There are 9 different objects: 8 men m m m m m m m m,  and 1 "4-women" object {wwww} arrange in 9! ways then for each arrange the 4 women in 4! ways.

7. C(14,6) - C(12,4)

There are C(14,6) total guest lists, subtract the C(12,4) ways that both feuding friends are invited.

8. C(16,8)

The path can be defined by the instructions of a 16 letter string of 8 e's (east) and 8 s's (south), for example eeeeeeessssssss. There are C(16,8) arrangements of these letters and thus the same number of east-south paths from corner to corner.

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C(n,r) = n!/(r!(n-r)!) = P(n,r)/r!

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Need help with some permutation and combination problems.?

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