MATH SOLVE

5 months ago

Q:
# A city distributes vehicle identification stickers using a combination of letters followed by a combination of digits, each of which may be used more than once. determine the number of possible stickers under the given circumstances. a) using 44 letters and 22 digits, how many stickers are possible? b) using 22 letters and 44 digits, how many stickers are possible? c) adding 1 letterletter before the 44 letters and 22 digits, how many stickers are possible?

Accepted Solution

A:

A) 672750

B) 68250

C) 2960100

Explanation

A) We have 26 letters from which to choose 4, and 10 digits from which to choose 2:

[tex]_{26}C_4\times_{10}C_2=\frac{26!}{4!22!}\times\frac{10!}{2!8!}=14950\times45=672750[/tex]

B) We have 26 letters from which to choose 2, and 10 digits from which to choose 4:

[tex]_{26}C_2\times_{10}C_4=\frac{26!}{2!24!}\times\frac{10!}{4!6!}=325\times210=68250[/tex]

C) We have 26 letters from which to choose 5, and 10 digits from which to choose 2:

[tex]_{26}C_5\times_{10}C_2=\frac{26!}{5!21!}\times\frac{10!}{2!8!}=65780\times45=2960100[/tex]

B) 68250

C) 2960100

Explanation

A) We have 26 letters from which to choose 4, and 10 digits from which to choose 2:

[tex]_{26}C_4\times_{10}C_2=\frac{26!}{4!22!}\times\frac{10!}{2!8!}=14950\times45=672750[/tex]

B) We have 26 letters from which to choose 2, and 10 digits from which to choose 4:

[tex]_{26}C_2\times_{10}C_4=\frac{26!}{2!24!}\times\frac{10!}{4!6!}=325\times210=68250[/tex]

C) We have 26 letters from which to choose 5, and 10 digits from which to choose 2:

[tex]_{26}C_5\times_{10}C_2=\frac{26!}{5!21!}\times\frac{10!}{2!8!}=65780\times45=2960100[/tex]