How many subsets of the above set contain the letter a but not the

Question

{a, b, c, d, e, f, g, h, i, j}

How many subsets of the above set contain the letter a but not the letters c or d?

A. 112
B. 120
C. 128
D. 248
E. 256

sujoykrdatta · Answer

With a included, atleast one or none of the other 7 (,excluding c and d) can be selected in 2^7 = 128 ways

Answer C

stne · Answer

Mr Dutta, has already provided the professional approach. Here is just an alternate expanded form of the above.

Note: subsets cannot contain  c  or  d , hence out of  10 , we have only  8  options to choose from.

But after choosing  a,  we have only  7 , options to choose from.

1  element subset  =  { a }  = 1

2  element subset  = { a,7_C_1 }  =1* 7

3  element subset  = { a, 7_C_2 }  =1*21

4  element subset  =  { a, 7_C_3 }  =1*35

5  element subset  =  { a, 7_C_4 }  =1*35

6  element subset  =  { a, 7_C_5 }  = 1*21

7  element subset  =  { a, 7_C_6 }  = 1*7

8  element subset  =  { a, 7_C_7 }  = 1*1

Total  = 1+7+21+35 +35 +21+7+1= 128

Ans C

Hope it helped.

MartyMurray · Answer

{a, b, c, d, e, f, g, h, i, j}

How many subsets of the above set contain the letter a but not the letters c or d?

Cases Approach

Count the number of letters other than a, c, or d:

7

So, including the subset of only a, there are 8 different sizes of subsets we can use.

Find the number of possible subsets of each size:

Only a ---> 1 way to choose

a + 1 other letter --> 7C1 = 7 ways to choose

a + 2 other letters -> 7C2 = 21 ways to choose

a + 3 other letters -> 7C3 = 35 ways to choose

a + 4 other letters -> 7C4 = 35 ways to choose

a + 5 other letters -> 7C5 = 21 ways to choose

a + 6 other letters -> 7C6 = 7 ways to choose

a + 7 other letters -> 7C7 = 1 way to choose

Add the cases:

1 + 7 + 21 + 35 + 35 + 21 + 7 + 1 = 128

Elements In or Out Approach

Count the number of possible elements other than a:

7

Determine the number of possibilities for each element:

Each element can be chosen, in, or not chosen, out.

So, there are 2 possibilities for each element.

In other words, we have the following.

a in ---> 1 possibility

b in or out -> 2 possibilities

e in or out -> 2 possibilities

f in or out -> 2 possibilities

g in or out -> 2 possibilities

h in or out -> 2 possibilities

i in or out -> 2 possibilities

j in or out -> 2 possibilities

Multiply the possibilities together:

1 × 2^7 = 128

A. 112
B. 120
C. 128
D. 248
E. 256

Correct answer:   C

How many subsets of the above set contain the letter a but not the

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How many subsets of the above set contain the letter a but not the

Prep Toolkit

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