Binomial Theorem Question 12
Question 12 - 2024 (31 Jan Shift 2)
If for some $\mathrm{m}, \mathrm{n} ;{ }^{6} \mathrm{C}{\mathrm{m}}+2\left({ }^{6} \mathrm{C}{\mathrm{m}+1}\right)+{ }^{6} \mathrm{C}{\mathrm{m}+2}>{ }^{8} \mathrm{C}{3}$ and ${ }^{n-1} P_{3}:{ }^{n} P_{4}=1: 8$, then ${ }^{n} P_{m+1}+{ }^{n+1} C_{m}$ is equal to
(1) 380
(2) 376
(3) 384
(4) 372
Show Answer
Answer (4)
Solution
${ }^{6} \mathrm{C}{\mathrm{m}}+2\left({ }^{6} \mathrm{C}{\mathrm{m}+1}\right)+{ }^{6} \mathrm{C}{\mathrm{m}+2}>{ }^{8} \mathrm{C}{3}$
${ }^{7} \mathrm{C}{\mathrm{m}+1}+{ }^{7} \mathrm{C}{\mathrm{m}+2}>{ }^{8} \mathrm{C}_{3}$
${ }^{8} \mathrm{C}{\mathrm{m}+2}>{ }^{8} \mathrm{C}{3}$
$\therefore \mathrm{m}=2$
And ${ }^{\mathrm{n}-1} \mathrm{P}{3}:{ }^{n} \mathrm{P}{4}=1: 8$
$\frac{(n-1)(n-2)(n-3)}{n(n-1)(n-2)(n-3)}=\frac{1}{8}$
$\therefore \mathrm{n}=8$
$\therefore{ }^{\mathrm{n}} \mathrm{P}{\mathrm{m}+1}+{ }^{\mathrm{n}+1} \mathrm{C}{\mathrm{m}}={ }^{8} \mathrm{P}{3}+{ }^{9} \mathrm{C}{2}$
$=8 \times 7 \times 6+\frac{9 \times 8}{2}$
$=372$
