Question

Lottery codes in the format XYZ are to be distributed. If X is an uppercase vowel, Y is an uppercase consonant, and Z can be any single-digit number, including 0, how many lottery codes are possible?

Answer 1

depends if you cound y as a vowel or not

if you don't count it as a vowel then
aeiou, 5

5*21*10=1050 codes



if you count it as a vowel
6*20*10=1200 codes

Answer 2

Answer:

1050

Step-by-step explanation:

Lottery codes in the format XYZ

X is an uppercase vowel

Y is an uppercase consonant

Z can be any single-digit number

So, there are 5 vowels.

We need to select 1 vowel from these 5

There 21 consonants.

We need to select 1 consonant from these 21

There are 10 single digit number 9(including 0)

We need to select 1 single digit number  from these 10

Now we will use combination to find  how many lottery codes are possible

Formula : $^nC_r= \frac{n!}{r!(n-r)!}$

No. of lottery tickets are possible:

=  $^5C_1\times ^{21}C_1 \times ^{10}C_1/tex]

=  [tex]\frac{5!}{1!(5-1)!}\times\frac{21!}{1!(21-1)!} \times\frac{10!}{1!(10-1)!}$

=  $\frac{5!}{1!(4)!}\times\frac{21!}{1!(20)!} \times\frac{10!}{1!(9)!}$

=  $\frac{5\times4!}{1!(4)!}\times\frac{21\times20!}{1!(20)!} \times\frac{10\times9!}{1!(9)!}$

=  $5\times21\times10$

=  $1050$

Hence 1050 lottery tickets are possible.