Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

John, Jim, Jay, and Jack have formed a band consisting of 4 instruments. If each of the boys can play all 4 instruments, how many different arrangements are possible? What if John and Jim can play all 4 instruments, but Jay and Jack can each play only piano and drums?

Short Answer

Expert verified

The different ways of arrangements are 24.

The ways of arrangement when John and Jim can play all 4 instruments, but Jay and Jack can each play only piano and drums are 4.

Step by step solution

01

.Given information  

John, Jim, Jay, and Jack k, each of the boys can play all 4 instruments

We have to find out the number of arrangements if four of them playing different instruments and when John and Jim can play all 4 instruments, but Jay and Jack can each play only piano and drums.

02

. Description of possible arrangements

If each of the boys plays all instruments, then using different permutations for 4 different instruments, we get there are 4!=4×3×2×1=24

different arrangements are possible.

Since Jay and jack play only piano and drums, then arrangements for them is 2! using different permutations concepts,

John and Jim can play all, but only 2 instruments are left there because of the other two people, hence the arrangement for them is 2!.

Thus possible arrangement when John and Jim can play all 4 instruments, but Jay and Jack can each play only piano and drums is 2!×2!=4

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In a certain community, there are 3 families consisting of a single parent and 1 child, 3 families consisting of a single parent and 2 children, 5 families consisting of 2 parents and a single child, 7 families consisting of 2 parents and 2 children, and 6 families consisting of 2 parents and 3 children. If a parent and child from the same family are to be chosen, how many possible choices are there?

If 12people are to be divided into 3committees of respective sizes 3,4,and 5,how many divisions are possible?

Two experiments are to be performed. The first can result in any one of m possible outcomes. If the first experiment results in an outcomei,then the second experiment can result in any ofni the possible outcomes, i = 1, 2, ..., m. What is the number of possible outcomes of the two experiments?

Ten weight lifters are competing in a team weight-lifting contest. The lifters 3are from the United States,4are from Russia, 2are from China, and 1are from Canada. If the scoring takes account of the countries that the lifters represent, but not their individual identities, how many different outcomes are possible from the point of view of scores? How many different outcomes correspond to results in which the United States has 1competitors in the top three and 2in the bottom three?

Let Hk(n)be the number of vectors x1,...,xkfor which each xiis a positive integer satisfying 1xinand x1x2,,xk.

(a)Without any computations, argue that

localid="1648218400232" H1(n)=nHk(n)=j=1nHk-1(j)k>1

Hint: How many vectors are there in which xk=j?

(b) Use the preceding recursion to compute H3(5).

Hint: First compute H2(n)forn=1,2,3,4,5.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free