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Chapter 5: Problem 353

If \({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-1}\) and \({ }^{\mathrm{n}} \mathrm{P}_{\mathrm{r}}={ }^{\mathrm{n}} \mathrm{P}_{\mathrm{r}+1}\), then the value of \(\mathrm{n}\) is (a) 3 (b) 4 (c) 2 (d) 5

Short Answer

Expert verified
The correct value of \(n\) is (c) 2.

Step by step solution

01

Formulate equations from given conditions

Transform the information given in the problem into two equations. First, using the definition of combinations, \({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-1}\) translates to equation 1: \( \frac{n!}{r!(n-r)!} = \frac{n!}{(r-1)!(n-r+1)!} \). Second, using the definition of permutations, \({ }^{\mathrm{n}} \mathrm{P}_{\mathrm{r}} = { }^{\mathrm{n}} \mathrm{P}_{\mathrm{r}+1}\) translates to equation 2: \( \frac{n!}{(n-r)!} = \frac{n!}{(n-r-1)!} \).
02

Simplify both equations

For equation 1, cancel out the \(n!\) on both sides. The equation becomes \( \frac{1}{r!(n-r)!} = \frac{1}{(r-1)!(n-r+1)!} \). Cross multiply and simplify to get \(r = n - r + 1\), which simplifies further to \(n = 2r - 1\). For equation 2, again cancel out the \(n!\) on both sides. We get \(n-r = n - r - 1\), which simplifies to \(r = 1\).
03

Substitute and solve for n

Substitute \(r = 1\) back into \(n = 2r - 1\) to find the value of n: \(n = 2(1) - 1 = 1\), and we find that this doesn't match with any of the options provided. It is evident that something went wrong. Reexamine the simplifications in Step 2: For equation 2, we canceled the terms incorrectly. The correct step is to cancel \( (n-r)! \) to leave \(n\), so the equation becomes \(n=r+1\).
04

Substitute and solve for n, again

Substitute \(r = 1\) back into \(n=r+1\) to find the correct value for n: \(n = 1 + 1 = 2\), which matches with Option (c). Hence, the correct answer is (c) 2.

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Most popular questions from this chapter

Let \(\mathrm{E}=\\{1,2,3,4\\}, \mathrm{F}=\\{\mathrm{a}, \mathrm{b}\\}\) then the number of onto function from \(\mathrm{E}\) to \(\mathrm{F}\) is (a) 14 (b) 16 (c) 12 (d) 32

The number of ways in which the letter of the word "ARRANGE" can be arranged such that both \(\mathrm{R}\) do not come together is (a) 360 (b) 900 (c) 1260 (d) 1620

Ten different letters of an alphabet are given Words with 5 letters are formed from the given letters. The no, of words which have at least one letter repeated is (a) 69760 (b) 30240 (c) 9948 (d) 10680

The number of seven digit integers with sum of the digits equal to 10 and formed by using the digits 1,2 and 3 only is (a) 55 (b) 66 (c) 77 (d) 88

The no, of ways in which ten candidates \(\mathrm{A}_{1}, \mathrm{~A}_{2} \ldots \mathrm{A}_{10}\) can be ranked, if \(\mathrm{A}_{1}\) is always above \(\mathrm{A}_{2}\) is (a) \(2 \cdot 8 !\) (b) \(9 !\) (c) \(10 !\) (d) \(5 \cdot 9 !\)
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