Acustomer ordered fourteen zingers. zingers are placed in packages of four, three, or one. in how many different ways can this order be filled?

fourteen zingers. zingers are placed in packages of four, three, or one

4*3=12

1*2=2

4*2=8

3*2=6

4*2=8

1*6=6

4*2=8

3*1=3

1*3=3

4*1=4

3*3=9

1*1=1

4*1=4

3*2=6

1*4=4

4*1=4

3*1=3

1*7=7

4*1=4

1*10=10

3*4=12

1*2=2

3*3=9

1*5=5

3*2=6

1*8=8

3*1=3

1*11=11

1*14=14

13 ways

The number of different ways this order can be filled is:

                              13 ways.

A customer ordered fourteen zingers.

zingers are placed in packages of​ four, three, or one.

Case-1

If the number of packages of four are: 3

Then 4×3=12 zingers.

a)

So, there will be zero packet of 3.

and 2 packs of 1.

Since, 12+2×1=14

Case-2

If the number of packages of four are: 2

i.e. 4×2= 8 zingers.

a)

If number of packages of 3 are: 2

i.e. 3×2=6

then number of package of 1 have to be 0.

b)

If number of packages of 3 are: 1

i.e. 3×1=3 zingers

Number of packages of 1 zinger will be: 3

i.e. 8+3+3=14

c)

If number of packages of 3 are: 0

Then number of packets of 1 will be:  6

Case-3

If the number of packages of four are: 1

a)

If number of packages of 3 are: 3

i.e. 3×3=9 zingers.

Hence, number of packets of 1 have to be: 1

b)

If number of packages of 3 are: 2

i.e. 3×2=6 zingers

Then number of packets of 1 have to be: 4

c)

If number of packages of 3 are: 1

i.e. 3×1=3 zingers

Then number of packets of 1 have to be: 7

d)

If number of packages of 3 are: 0

i.e. 3×0=0 zingers

Then number of packets of 1 have to be: 10

Case-4

If the number of packages of four are: 0

a)

If number of packages of 3 are: 4

i.e. 3×4=12 zingers

Then number of packets of 1 have to be: 2

b)

If number of packages of 3 are: 3

i.e. 3×3=9 zingers

Then number of packets of 1 have to be: 5

c)

If number of packages of 3 are: 2

i.e. 3×2=6 zingers

Then number of packets of 1 have to be: 8

d)

If number of packages of 3 are: 1

i.e. 3×1=3 zingers

Then number of packets of 1 have to be: 11

e)

If number of packages of 3 are: 0

i.e. 3×0=0 zingers

Then number of packets of 1 have to be: 14

There are total 13 ways of doing so.

( 1 from case-1

3 from case-2

4 from case-3

and 5 from case-4

i.e.  1+3+4+5=13 ways )

answer:  

123455421

step-by-step explanation: easy

.632

step-by-step explanation:

6 * 1/10 = 3/5

3*0.01 = 0.03

3/5 + 0.03 = .63

2*0.002 = .002

.63 + 0.002 = .632