Question

how many ways can 3 students be arraanged in 5 benches

Answer 1

Here, arrangement is occur. So the answer is 5P3.

$^5P_3 = \frac{5!}{(5 - 3)!} = \frac{5!}{2!} = \frac{5 \times 4 \times 3 \times 2!}{2!} = 5 \times 4 \times 3 = 60$

∴ 3 students can be arranged in 5 benches in 60 different ways.

Let's check.

Let the benches be indicated with letters A, B, C, D and E.

Given below are the ways in which 3 students can be arranged:

ABC, ABD, ABE, ACB, ACD, ACE,
ADB, ADC, ADE, AEB, AEC, AED,
BAC, BAD, BAE, BCA, BCD, BCE,
BDA, BDC, BDE, BEA, BEC, BED,
CAB, CAD, CAE, CBA, CBD, CBE,
CDA, CDB, CDE, CEA, CEB, CED,
DAB, DAC, DAE, DBA, DBC, DBE,
DCA, DCB, DCE, DEA, DEB, DEC,
EAB, EAC, EAD, EBA, EBC, EBD,
ECA, ECB, ECD, EDA, EDB, EDC.

So there are 60 ways.

Don't think that same arrangements have been occur several times.

E. g.: There are 6 arrangements ABC, ACB, BAC, BCA, CAB and CBA. These 6 arrangements are not same as the 3 students are arranged differ from each arrangement. If the students are indicated by number 1, 2 and 3, you can understand it.

In ABC and CAB, 1st student is at A in first arrangement but at C in second arrangement. The same in 2nd and 3rd students too.

Hope my answer can be understood by you.

Please mark my answer as the brainliest if this may be helpful.

Thank you. Have a nice day.

#adithyasajeevan