Chapter 1: Q. 1.18 (page 19)

Argue that
$(\begin{matrix} n \\ n_{1}, n_{2}, . . . . . . . ., n_{r} \end{matrix}) = (\begin{matrix} n - 1 \\ n_{1} - 1, n_{2}, . . . . . . . ., n_{r} \end{matrix}) + (\begin{matrix} n - 1 \\ n_{1}, n_{2} - 1, . . . . . . . ., n_{r} \end{matrix}) + . . . . . . + (\begin{matrix} n - 1 \\ n_{1}, n_{2}, . . . . . . . ., n_{r} - 1 \end{matrix})$
Hint: Use an argument similar to the one used to establish Equation (4.1).

Short Answer

Expert verified

It is proved that

$(\begin{matrix} n \\ n_{1}, n_{2}, . . . . . . . ., n_{r} \end{matrix}) = (\begin{matrix} n - 1 \\ n_{1} - 1, n_{2}, . . . . . . . ., n_{r} \end{matrix}) + (\begin{matrix} n - 1 \\ n_{1}, n_{2} - 1, . . . . . . . ., n_{r} \end{matrix}) + . . . . . . + (\begin{matrix} n - 1 \\ n_{1}, n_{2}, . . . . . . . ., n_{r} - 1 \end{matrix})$

Step by step solution

Step 1. Given information.

We have to prove that $(\begin{matrix} n \\ n_{1}, n_{2}, . . . . . . . ., n_{r} \end{matrix}) = (\begin{matrix} n - 1 \\ n_{1} - 1, n_{2}, . . . . . . . ., n_{r} \end{matrix}) + (\begin{matrix} n - 1 \\ n_{1}, n_{2} - 1, . . . . . . . ., n_{r} \end{matrix}) + . . . . . . + (\begin{matrix} n - 1 \\ n_{1}, n_{2}, . . . . . . . ., n_{r} - 1 \end{matrix})$

Step 2. Prove that nn1, n2, ........ , nr=n-1n1-1, n2, ........ , nr+n-1n1, n2-1, ........ , nr+......+n-1n1, n2, ........ , nr-1

$(\begin{matrix} n \\ n_{1}, n_{2}, . . . . . . . ., n_{r} \end{matrix})$ represents the no. of possible divisions of $n$ distinct objects into $r$ distinct groups of respective sizes $n_{1}, n_{2}, . . . . . . . . ., n_{r}$ .

Now, let's concentrate on one object, suppose it is grouped in $n_{1}$ group, it can be done in $(\begin{matrix} n - 1 \\ n_{1} - 1, n_{2}, . . . . . . . ., n_{r} \end{matrix})$ ways.

If it is not grouped in $n_{1}$ group but it is grouped in $n_{2}$ group, it can be done in $(\begin{matrix} n - 1 \\ n_{1}, n_{2} - 1, . . . . . . . ., n_{r} \end{matrix})$ ways.

Similarly, if it is grouped in $n_{r}$ group, it can be done in $(\begin{matrix} n - 1 \\ n_{1}, n_{2}, . . . . . . . ., n_{r} - 1 \end{matrix})$ ways.

Hence, it is proved that $(\begin{matrix} n \\ n_{1}, n_{2}, . . . . . . . ., n_{r} \end{matrix}) = (\begin{matrix} n - 1 \\ n_{1} - 1, n_{2}, . . . . . . . ., n_{r} \end{matrix}) + (\begin{matrix} n - 1 \\ n_{1}, n_{2} - 1, . . . . . . . ., n_{r} \end{matrix}) + . . . . . . + (\begin{matrix} n - 1 \\ n_{1}, n_{2}, . . . . . . . ., n_{r} - 1 \end{matrix})$

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Short Answer

Step by step solution

Step 1. Given information.

Step 2. Prove that nn1, n2, ........ , nr=n-1n1-1, n2, ........ , nr+n-1n1, n2-1, ........ , nr+......+n-1n1, n2, ........ , nr-1

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Argue thatnn1,n2,........,nr=n-1n1-1,n2,........,nr+n-1n1,n2-1,........,nr+......+n-1n1,n2,........,nr-1Hint: Use an argument similar to the one used to establish Equation (4.1).

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Prove that nn1, n2, ........ , nr=n-1n1-1, n2, ........ , nr+n-1n1, n2-1, ........ , nr+......+n-1n1, n2, ........ , nr-1

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Statistics

Calculus

Probability and Statistics

Mechanics Maths

Decision Maths

Pure Maths

Study anywhere. Anytime. Across all devices.