Question

problem a
8 books are to be arranged on a shelf. in how many ways can this be done if 3 particular books must be kept together?

Explanation:

Step1: Treat the 3 books as a single unit

When 3 particular books must be kept together, we can consider them as one combined unit. So now, instead of 8 individual books, we have \(8 - 3+ 1=6\) units to arrange (the group of 3 books and the remaining 5 individual books). The number of ways to arrange these 6 units is given by the permutation formula for \(n\) distinct objects, which is \(n!\). So the number of ways to arrange 6 units is \(6!\).

Step2: Arrange the 3 books within the group

The 3 particular books within the group can be arranged among themselves. The number of permutations of 3 distinct books is \(3!\) (using the permutation formula \(P(n,r)=\frac{n!}{(n - r)!}\), here \(n = 3\) and \(r=3\), so \(P(3,3)=\frac{3!}{(3 - 3)!}=3!\)).

Step3: Calculate the total number of arrangements

To find the total number of ways to arrange the 8 books with the 3 particular books together, we multiply the number of ways to arrange the 6 units (the group and the other books) by the number of ways to arrange the 3 books within the group. So the total number of arrangements is \(6!\times3!\).

We know that \(n!=n\times(n - 1)\times\cdots\times1\), so \(6!=6\times5\times4\times3\times2\times1 = 720\) and \(3!=3\times2\times1=6\). Then \(6!\times3!=720\times6 = 4320\).

Answer:

The number of ways to arrange the 8 books with 3 particular books kept together is \(\boldsymbol{4320}\).