Question

$\huge\underline{\pink{\bf{Question:-}}}$

=> ꜰɪɴᴅ ᴛʜᴇ ɴᴜᴍʙᴇʀ ᴏꜰ ꜱᴜʙꜱᴇᴛꜱ ᴏꜰ ᴀ×ʙ ᴡʜᴇʀᴇ ᴀ ={3,6,9} ᴀɴᴅ ʙ={x,ʏ,ᴢ}

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Answer 1

Answer:

What are all of the subsets of the set (3, 6, 9, 12)?

So, first off, for any set with number of elements n there are 2^n subsets of it. This is useful to check at the end that you didn't miss any. In this case that is 2^4=16 subsets.

Try to do it in a systematic way. Here's how I do it.

First, all sets have the subset {}, aka the empty set.

Then there are {3}, {6}, {9}, {12}

Then there is {3, 6}, {3, 9}, {3, 12}

Then {6, 9}, {6, 12}, and {9, 12}

Then {3, 6, 9}, { 3, 6, 12}, { 3, 9, 12}, {6, 9, 12}

Lastly every set is a subset of itself, so {3, 6, 9, 12}

Now I count them up, see that I listed 16 sets, check and see that there are no duplicates, and so I'm done.

Explanation:

#Hope you have satisfied with this answer.

Answer 2

Answer:

What are all of the subsets of the set (3, 6, 9, 12)?

So, first off, for any set with number of elements n there are 2^n subsets of it. This is useful to check at the end that you didn't miss any. In this case that is 2^4=16 subsets.

Try to do it in a systematic way. Here's how I do it.

First, all sets have the subset {}, aka the empty set.

Then there are {3}, {6}, {9}, {12}

Then there is {3, 6}, {3, 9}, {3, 12}

Then {6, 9}, {6, 12}, and {9, 12}

Then {3, 6, 9}, { 3, 6, 12}, { 3, 9, 12}, {6, 9, 12}

Lastly every set is a subset of itself, so {3, 6, 9, 12}

Now I count them up, see that I listed 16 sets, check and see that there are no duplicates, and so I'm done.