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2 years ago

in how many ways can 4 mangoes and 3 apples be distributed among 7 children if each child receives only 2 fruit?

Mathematics

1 answer:

abruzzese [7]2 years ago

7 0

solution

this is what I did. total number of ways without any restriction =900. we have to subtract the case when at least one child doesn't get any fruit. let that be C1∪C2∪C3 and we use the principle of mutual inclusion and exclusion to find that. C1∪C2∪C3=C1+C2+C3−(all two intersections)+3 intersections (no one gets a fruit case) =3⋅60−3+1=178 so case where each child gets at least one fruit =900−178=722.

Is this right, someone saying answer =49

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Answer:

Step-by-step explanation:

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2 years ago

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Step-by-step explanation:

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2 years ago

There are eight different jobs in a printer queue. Each job has a distinct tag which is a string of three upper case letters. Th

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Answer:

a. 40320 ways

b. 10080 ways

c. 25200 ways

d. 10080 ways

e. 10080 ways

Step-by-step explanation:

There are 8 different jobs in a printer queue.

a. They can be arranged in the queue in 8! ways.

No. of ways to arrange the 8 jobs = 8!

= 8*7*6*5*4*3*2*1

No. of ways to arrange the 8 jobs = 40320 ways

b. USU comes immediately before CDP. This means that these two jobs must be one after the other. They can be arranged in 2! ways. Consider both of them as one unit. The remaining 6 together with both these jobs can be arranged in 7! ways. So,

No. of ways to arrange the 8 jobs if USU comes immediately before CDP

= 2! * 7!

= 2*1 * 7*6*5*4*3*2*1

= 10080 ways

c. First consider a gap of 1 space between the two jobs USU and CDP. One case can be that USU comes at the first place and CDP at the third place. The remaining 6 jobs can be arranged in 6! ways. Another case can be when USU comes at the second place and CDP at the fourth. This will go on until CDP is at the last place. So, we will have 5 such cases.

The no. of ways USU and CDP can be arranged with a gap of one space is:

6! * 6 = 4320

Then, with a gap of two spaces, USU can come at the first place and CDP at the fourth. This will go on until CDP is at the last place and USU at the sixth. So there will be 5 cases. No. of ways the rest of the jobs can be arranged is 6! and the total no. of ways in which USU and CDP can be arranged with a space of two is: 5 * 6! = 3600

Then, with a gap of three spaces, USU will come at the first place and CDP at the fifth. We will have four such cases until CDP comes last. So, total no of ways to arrange the jobs with USU and CDP three spaces apart = 4 * 6!

Then, with a gap of four spaces, USU will come at the first place and CDP at the sixth. We will have three such cases until CDP comes last. So, total no of ways to arrange the jobs with USU and CDP three spaces apart = 3 * 6!

Then, with a gap of five spaces, USU will come at the first place and CDP at the seventh. We will have two such cases until CDP comes last. So, total no of ways to arrange the jobs with USU and CDP three spaces apart = 2 * 6!

Finally, with a gap of 6 spaces, USU at first place and CDP at the last, we can arrange the rest of the jobs in 6! ways.

So, total no. of different ways to arrange the jobs such that USU comes before CDP = 10080 + 6*6! + 5*6! + 4*6! + 3*6! + 2*6! + 1*6!

= 10080 + 4320 + 3600 + 2880 + 2160 + 1440 + 720

= 25200 ways

d. If QKJ comes last then, the remaining 7 jobs can be arranged in 7! ways. Similarly, if LPW comes last, the remaining 7 jobs can be arranged in 7! ways. so, total no. of different ways in which the eight jobs can be arranged is 7! + 7! = 10080 ways

e. If QKJ comes last then, the remaining 7 jobs can be arranged in 7! ways in the queue. Similarly, if QKJ comes second-to-last then also the jobs can be arranged in the queue in 7! ways. So, total no. of ways to arrange the jobs in the queue is 7! + 7! = 10080 ways

3 0

2 years ago

in how many ways can 4 mangoes and 3 apples be distributed among 7 children if each child receives only 2 fruit?​

in how many ways can 4 mangoes and 3 apples be distributed among 7 children if each child receives only 2 fruit?