Question

Column I		Column II
A	Number of subsets of $\{a,b,c,d,e,f,g\}$ which contain both $a$ and $b$, are	P	120
B	Number of distinguishable arrangements of the letters in CONTEST in which first two places are occupied by vowels, are	Q	240
C	Number of permutations of the letters ABCDEFG which contain the word $BAD$, are	R	60
D	Nine chairs in a row are to be occupied by six students and Professors Alpha, Beta and Gamma. The three professors arrive before the six students and decide to choose their chairs so that each professor will be between two students. Number of ways can professors Alpha, Beta, and Gamma can sit on their chairs, are	S	32

• A. $(A)\to S,(B)\to P,(C)\to Q,(D)\to R$
• B. $(A)\to S,(B)\to P,(C)\to Q,(D)\to P$
• C. $(A)\to S,(B)\to P,(C)\to Q,(D)\to Q$
• D. $(A)\to S,(B)\to P,(C)\to Q,(D)\to S$

Answer 1

Answer: (A)

(A) $\{a,b,c,d,e,f,g\}$
no. of subsets which contain a & $b={}^5{C}_1+{}^5{C}_2+{}^5{C}_3+{}^5{C}_4+{}^5{C}_5{}^{+5}{C}_0=2^5=32$
(B) $2!\times \frac{5!}{2!}=120$
$\left[12^{\text{th }}(14-5-2006)\right]$
(C) Let us denote by $W$ the word BAD. One possible permutation that contains the word $W=BAD$ is WCEFG. This means that in order to count all the permutations with the word BAD in them we need to count all the permutations of the letter W, C, E, F, and G. This count
yields $5!=120$
(D) $C_1{C}_2\dots \dots ..C_7{C}_8{C}_9$
$\therefore$ Each professor must be seated between two students hence the professor can take their chairs from $C_2,C_3\dots \dots C_7{C}_8$,
so that number of selection ${}^5{C}_3=10$
Total ways $=10\cdot 3!=60$