# If there are 6 periods in each working day of a school, then the number of ways that you can arrange 5 subjects during the working day is

a) 600

b) 720

c) 1800

d) 1325

Answer

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Hint: First see how many subjects you are provided with and the no. of periods available for these subjects. Select the subject assigned two periods and then the select two periods which it is assigned. And then assign the other four subjects to four periods.

We have 5 subjects and periods, therefore it is very clear that once subject will occupy 2 periods every day

Let the 5 subjects be A, B, C, D, E and six periods be 1 to 6.

Subject a can have any of the six periods, similarly, B, C, D and E.

Subjects can be arranged in 6 periods in = \[{}^6{P_5}\]

Remaining 1 period can be arranged in = \[{}^5{P_1}\]

Two subjects are alike in each of the arrangement,

So we need to divide by 2! To avoid overcounting

Total no. of arrangements = \[\dfrac{{{}^6{P_5} \times {}^5{P_6}}}{{2!}}\]………………(1)

\[P(n,r) = \dfrac{{n!}}{{(n - r)!}}\]

P is no. of permutations

N is the total no. of objects in the set

r is the no. of choosing objects in the set

\[{}^6{P_5} = \dfrac{{6!}}{{(6 - 5)!}}\] = 720

\[{}^5{P_1} = \dfrac{{5!}}{{(5 - 1)!}}\] = 5

Putting these values in equation (1) we get

\[\dfrac{{720 \times 5}}{{2!}}\] = 1800

So the correct option is ‘c’.

Note: Permutation of a set is an arrangement of its elements into a sequence or linear order, or if it is already ordered, a rearrangement of it’s elements. In this question also we rearranged 5 subjects in 6 periods.

We have 5 subjects and periods, therefore it is very clear that once subject will occupy 2 periods every day

Let the 5 subjects be A, B, C, D, E and six periods be 1 to 6.

Subject a can have any of the six periods, similarly, B, C, D and E.

Subjects can be arranged in 6 periods in = \[{}^6{P_5}\]

Remaining 1 period can be arranged in = \[{}^5{P_1}\]

Two subjects are alike in each of the arrangement,

So we need to divide by 2! To avoid overcounting

Total no. of arrangements = \[\dfrac{{{}^6{P_5} \times {}^5{P_6}}}{{2!}}\]………………(1)

\[P(n,r) = \dfrac{{n!}}{{(n - r)!}}\]

P is no. of permutations

N is the total no. of objects in the set

r is the no. of choosing objects in the set

\[{}^6{P_5} = \dfrac{{6!}}{{(6 - 5)!}}\] = 720

\[{}^5{P_1} = \dfrac{{5!}}{{(5 - 1)!}}\] = 5

Putting these values in equation (1) we get

\[\dfrac{{720 \times 5}}{{2!}}\] = 1800

So the correct option is ‘c’.

Note: Permutation of a set is an arrangement of its elements into a sequence or linear order, or if it is already ordered, a rearrangement of it’s elements. In this question also we rearranged 5 subjects in 6 periods.

Last updated date: 28th Sep 2023

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