118919
If \(8{ }^{\prime} P_r=7 .{ }^8 P_{r-1}\) what is the value of ' \(r\) '?
1 \(3 \& 4\)
2 \(2 \& 5\)
3 4 \& 5
4 No solution (No such 'r' exists)
Explanation:
D : Given, \({ }^7 \mathrm{P}_{\mathrm{r}}=7 .{ }^8 \mathrm{P}_{\mathrm{r}-1}\) We know that, \({ }^n P_r=\frac{n !}{(n-r) !}\) \(\therefore \quad 8 \frac{7 !}{(7-r) !}=\frac{7 \times 8 !}{(9-r) !}=\frac{8 \times 7 \times 7 !}{(9-r) !}\) \(\frac{1}{(7-r) !}=\frac{7}{(9-r) !}\) Therefore \(\mathrm{r}\) should be an integer, but no two integers can multiply to give 7 to \(\mathrm{r}\), which has no solutions. So no such \(\mathrm{r}\) exists.
118919
If \(8{ }^{\prime} P_r=7 .{ }^8 P_{r-1}\) what is the value of ' \(r\) '?
1 \(3 \& 4\)
2 \(2 \& 5\)
3 4 \& 5
4 No solution (No such 'r' exists)
Explanation:
D : Given, \({ }^7 \mathrm{P}_{\mathrm{r}}=7 .{ }^8 \mathrm{P}_{\mathrm{r}-1}\) We know that, \({ }^n P_r=\frac{n !}{(n-r) !}\) \(\therefore \quad 8 \frac{7 !}{(7-r) !}=\frac{7 \times 8 !}{(9-r) !}=\frac{8 \times 7 \times 7 !}{(9-r) !}\) \(\frac{1}{(7-r) !}=\frac{7}{(9-r) !}\) Therefore \(\mathrm{r}\) should be an integer, but no two integers can multiply to give 7 to \(\mathrm{r}\), which has no solutions. So no such \(\mathrm{r}\) exists.
118919
If \(8{ }^{\prime} P_r=7 .{ }^8 P_{r-1}\) what is the value of ' \(r\) '?
1 \(3 \& 4\)
2 \(2 \& 5\)
3 4 \& 5
4 No solution (No such 'r' exists)
Explanation:
D : Given, \({ }^7 \mathrm{P}_{\mathrm{r}}=7 .{ }^8 \mathrm{P}_{\mathrm{r}-1}\) We know that, \({ }^n P_r=\frac{n !}{(n-r) !}\) \(\therefore \quad 8 \frac{7 !}{(7-r) !}=\frac{7 \times 8 !}{(9-r) !}=\frac{8 \times 7 \times 7 !}{(9-r) !}\) \(\frac{1}{(7-r) !}=\frac{7}{(9-r) !}\) Therefore \(\mathrm{r}\) should be an integer, but no two integers can multiply to give 7 to \(\mathrm{r}\), which has no solutions. So no such \(\mathrm{r}\) exists.
118919
If \(8{ }^{\prime} P_r=7 .{ }^8 P_{r-1}\) what is the value of ' \(r\) '?
1 \(3 \& 4\)
2 \(2 \& 5\)
3 4 \& 5
4 No solution (No such 'r' exists)
Explanation:
D : Given, \({ }^7 \mathrm{P}_{\mathrm{r}}=7 .{ }^8 \mathrm{P}_{\mathrm{r}-1}\) We know that, \({ }^n P_r=\frac{n !}{(n-r) !}\) \(\therefore \quad 8 \frac{7 !}{(7-r) !}=\frac{7 \times 8 !}{(9-r) !}=\frac{8 \times 7 \times 7 !}{(9-r) !}\) \(\frac{1}{(7-r) !}=\frac{7}{(9-r) !}\) Therefore \(\mathrm{r}\) should be an integer, but no two integers can multiply to give 7 to \(\mathrm{r}\), which has no solutions. So no such \(\mathrm{r}\) exists.
118919
If \(8{ }^{\prime} P_r=7 .{ }^8 P_{r-1}\) what is the value of ' \(r\) '?
1 \(3 \& 4\)
2 \(2 \& 5\)
3 4 \& 5
4 No solution (No such 'r' exists)
Explanation:
D : Given, \({ }^7 \mathrm{P}_{\mathrm{r}}=7 .{ }^8 \mathrm{P}_{\mathrm{r}-1}\) We know that, \({ }^n P_r=\frac{n !}{(n-r) !}\) \(\therefore \quad 8 \frac{7 !}{(7-r) !}=\frac{7 \times 8 !}{(9-r) !}=\frac{8 \times 7 \times 7 !}{(9-r) !}\) \(\frac{1}{(7-r) !}=\frac{7}{(9-r) !}\) Therefore \(\mathrm{r}\) should be an integer, but no two integers can multiply to give 7 to \(\mathrm{r}\), which has no solutions. So no such \(\mathrm{r}\) exists.
