The elements of set S are all positive integers that can be expressed

Question

The elements of set S are all positive integers that can be expressed as the product of two or more of the factors 3, 11, 17 and 23. If no single product can use the same factor twice, how many elements are in Set S?

A 6
B 9
C 11
D 24
E 16

A 6
B 9
C 11
D 24
E 16

chandy123 · Answer

To determine the number of elements in set S, we need to find all possible products formed by multiplying two or more distinct factors from the set {3, 11, 17, 23}.

1. Identify the Number of Ways to Choose Factors:
 • Choosing 2 factors:
The number of ways to choose 2 distinct factors from 4 is calculated using combinations:
C(4, 2) = 6
These combinations are:
(3 × 11), (3 × 17), (3 × 23), (11 × 17), (11 × 23), (17 × 23)
 • Choosing 3 factors:
The number of ways to choose 3 distinct factors from 4:
C(4, 3) = 4
These combinations are:
(3 × 11 × 17), (3 × 11 × 23), (3 × 17 × 23), (11 × 17 × 23)
 • Choosing all 4 factors:
There’s only one way to choose all 4 factors:
C(4, 4) = 1
Combination:
(3 × 11 × 17 × 23)
 2. Calculate the Total Number of Elements:
 • Total combinations = C(4, 2) + C(4, 3) + C(4, 4)
 • Total combinations = 6 + 4 + 1 = 11
 3. Ensure All Products Are Unique:
 • Since all factors are distinct prime numbers, each product will be unique. There are no duplicate products.

Conclusion

There are 11 distinct elements in set S.
Option C is the correct answer.

MalachiKeti · Answer

S = {1*3,1*11,1*17,1*23}

Unique Factors are 1,3,11,17,23

Using 1 as factor = 1*3 | 11*1 | 17*1 | 23*1
Product of 2 factors   = 3*11 | 3* 17 | 3*23 | 11*17| 11*23 | 17*23 
 Product of 3 factors   = 3*11*17 | 3*11*23 | 3*17*23 | 11*23*17
 Product of 4 factors   = 3*11*17*23

_ _+ _ _ + _ _ _ + _ _ _ _
4 + 4*3/2! + 4*3*2/3! + 1

4 + 6 + 4 + 1 = 15

MalachiKeti · Answer

Bunuel  why is 1 avoided?

Bunuel · Answer

1 is not included because the question treats 3, 11, 17, and 23 as the factors themselves. It’s not about their individual factors, so 1 is irrelevant.

MalachiKeti · Answer

So the word 'Factors' is irrelevant right? It should simply say numbers 3,11,17,23. Even if we had numbers like 22,12,17 and 23 we would have gone with same approach?  Bunuel

Bunuel · Answer

Yes, here each of 3, 11, 17, and 23 is called factors.

Sujithz001 · Answer

Bunuel  Can this be solved through permutations & combinations? If yes can you provide a solution pls? If not, pls suggest the effective way to go about this problem.

Thanks.

Krunaal · Answer

We can solve this quickly if we think in terms of making subsets from the given set of {3, 11, 17, 23}

Total number of subsets =  2^{4} = 16

We will subtract 1 blank subset i.e. {} and 4 single subsets i.e. {3} {11} {17} {23}

16 - 1 - 4 =  11

Answer C.

sujoykrdatta · Answer

You can select 2 at a time in 4C2 = 6 ways
Or 3 at a time in 4C3 = 4 ways
Or 4 at a time in 4C4 = 1 way

Total = 6+4+1 = 11 ways

Ans C

essetempore · Answer

Another approach- simply just list out all the possible sets - {}, {3}, {11}, {17}, {23}, {3,11}, {3,17}, {3,23}, {11,17}, {11,23}, {17, 23}, {3,11,17}, {3, 11, 23}, {3, 17, 23}, {11, 17, 23}, and {3,11,17,23} 
Now cancel all the singly digit sets and the empty set, we will have 11

egmat · Answer

Yes, this is a pure combinations problem!

We have  4  factors: 3, 11, 17, and 23

We need to count how many different products we can make using  2 or more  factors (with no repetition).

Since multiplication is commutative (3 × 11 = 11 × 3), the ORDER doesn't matter. This means we use COMBINATIONS, not permutations.

The Solution:

Products using exactly  2  factors: C(4,2) =  6 
Products using exactly  3  factors: C(4,3) =  4 
Products using exactly  4  factors: C(4,4) =  1

Total =  6 + 4 + 1 = 11

Answer: C

Quick Verification: 
Products of 2: 3×11, 3×17, 3×23, 11×17, 11×23, 17×23 →  6  products
Products of 3: 3×11×17, 3×11×23, 3×17×23, 11×17×23 →  4  products
Products of 4: 3×11×17×23 →  1  product

Total:  11  elements in set S

Siddharth0007 · Answer

why a blank set to be subtracted ?

Bunuel · Answer

2^4 counts every way to choose from {3, 11, 17, 23}, including choosing nothing.

If you choose nothing ({}), you are not forming a product of two or more factors, so it cannot be in S. That is why the blank subset must be subtracted.

The 4 one element subsets also must be subtracted because they use only one factor, but S requires at least two.

The elements of set S are all positive integers that can be expressed

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The elements of set S are all positive integers that can be expressed

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