PERMS AND CONS Help please (1 Viewer)

[Footyking]

I am really struggling with these 3 permutation and combination questions and so i was wondering if anyone on here knows how to do them?


Q1: Find the number of ways in which 3 boys and 3 girls can be arranged in a straight line so that the 3 boys are not all next to each other.

Q2: Find the number of ways in which a group of 3 people can be chosen from 4 men and 5 women so that the group contains a majority of women.

Q3: A 3 letter code is to be made from the 26 letters of the alphabet (21 consonants and 5 vowels ) where repetition of letters is not allowed.
a. find the number of such three letter codes in which the middle letter is a vowel
b. find the number of three letter codes which contain 2 consonants and 1 vowel.

If anyone can help me with 1 of the questions or all of the questions it would be greatly appreciated

 

[jmk123]

I am really struggling with these 3 permutation and combination questions and so i was wondering if anyone on here knows how to do them?


Q1: Find the number of ways in which 3 boys and 3 girls can be arranged in a straight line so that the 3 boys are not all next to each other.
3 boys together are one group, arranged in 3! ways, times by 3! girls, so 4!x3!, then 6!-ans.

Q2: Find the number of ways in which a group of 3 people can be chosen from 4 men and 5 women so that the group contains a majority of women.
2 women or 3 women for majority. For 2 women, (5C2x4C1), for three women, (5C3x4C0). So just add those two cases.

Q3: A 3 letter code is to be made from the 26 letters of the alphabet (21 consonants and 5 vowels ) where repetition of letters is not allowed.
a. find the number of such three letter codes in which the middle letter is a vowel. You can choose 1 of 5 vowels for middle letter, so 5 choices. 21 letters remain for other two spaces, so 21P2. Add results.
b. find the number of three letter codes which contain 2 consonants and 1 vowel.
5 vowels in 3 positions can be, so 5^3=125 options for vowels. The consonants can be arranged in 21C2 ways again, but in three different ways, so (21C2)^3


If anyone can help me with 1 of the questions or all of the questions it would be greatly appreciated

Not 100 percent sure but I tried hehe.

 

[EpikHigh]

Q1) 3 boys not together is the complement of them all together, total re-arrangements = 6! = 720. Number of ways that boys are arranged being together = 3!x4! = 144. Therefore total ways that the boys can be arranged without having them all together = 720 - 144 = 576 ways

Only did question one because I believe the post above is wrong.

 

[jmk123]

Q1) 3 boys not together is the complement of them all together, total re-arrangements = 6! = 720. Number of ways that boys are arranged being together = 3!x4! = 144. Therefore total ways that the boys can be arranged without having them all together = 720 - 144 = 576 ways

Only did question one because I believe the post above is wrong.

Fixed it
I accidentally wrote 3!x3! instead of 4!x3!

OH WAIT ur right lol I thought it said "the boys ARE together"

indeed

6!-(4!x3!)

 

[Footyking]

Thanks heaps for the help. Really appreciate it