Seven cards, each marked a letter can be arranged to spell NUMBERS. how many three letter code words can be formed with 2 consonants and one vowel?

Thanks 😁

let's choose the consonants and vowels first

No of combinations
= C(5,2) x C(2,1) = 10(2) = 20

Since you titled your post "permutations" we now want to arrange these.
That can be done in 6 ways each

So the number of permutations is 6(20) = 120

To determine the number of three-letter code words that can be formed with 2 consonants and one vowel using the letters N, U, M, B, E, R, and S, we need to break down the problem into steps:

Step 1: Calculate the number of ways to choose 2 consonants from the given set.
Step 2: Calculate the number of ways to choose 1 vowel from the given set.
Step 3: Multiply the results of Step 1 and Step 2 together to get the total number of possible code words.

Now, let's start with Step 1:

There are 5 consonants in the set: N, M, B, R, and S

To choose 2 consonants from this set, we can use the combination formula:

nCr = n! / ((n-r)! * r!)

where n is the total number of items and r is the number of items being chosen.

In this case, n = 5 (the total number of consonants) and r = 2 (the number of consonants we want to choose).

Plugging in the values, we have:

5C2 = 5! / ((5-2)! * 2!)
= 5! / (3! * 2!)
= (5 * 4 * 3!) / (3! * 2 * 1)
= (5 * 4) / (2 * 1)
= 10

Therefore, there are 10 ways to choose 2 consonants from the given set.

Moving on to Step 2:

There are 2 vowels in the set: U and E

To choose 1 vowel from this set, we simply have 2 options (U or E).

Finally, for Step 3:

We multiply the results of Step 1 and Step 2 together:

Total number of three-letter code words = 10 (number of ways to choose 2 consonants) * 2 (number of ways to choose 1 vowel)
= 20

So, there are a total of 20 three-letter code words that can be formed with 2 consonants and one vowel using the letters N, U, M, B, E, R, and S.