132.
If all permutations of the letters of the word $$AGAIN$$ are arranged as in dictionary, then fiftieth word is
A
$$NAAGI$$
B
$$NAGAI$$
C
$$NAAIG$$
D
$$NAIAG$$
Answer :
$$NAAIG$$
Starting with the letter $$A$$ and arranging the other four letters, there are $$4! = 24$$ words. These are the first 24 words. Then starting with $$G$$ and arranging $$A, A, I$$ and $$N$$ in different ways, there are $$\frac{{4!}}{{2!1!1!}} = \frac{{24}}{2} = 12{\text{ words}}{\text{.}}$$
Hence, total 36 words.
Next, the $$37^{th}$$ word starts with $$I.$$ There are 12 words starting with $$I.$$ This accounts up to the $$48^{th}$$ word. The $$49^{th}$$ word is $$NAAGI.$$
The $$50^{th}$$ word is $$NAAIG.$$
133.
How many different words can be formed by jumbling the letters in the word $$MISSISSIPPI$$ in which no two $$S$$ are adjacent ?
First let us arrange $$M, I, I, I, I, P, P$$
Which can be done in $$\frac{{7!}}{{4!2!}}{\text{ways}}{\text{.}}$$
Now $$4\,S$$ can be kept at any of the ticked places in $$^8{C_4}$$ ways so that no two $$S$$ are adjacent.
Total required ways
$$ = \frac{{7!}}{{4!2!}}{\,^8}{C_4} = 7 \times {\,^6}{C_4} \times {\,^8}{C_4}$$
134.
The number of six digit numbers that can be formed from the digits 1, 2, 3, 4, 5, 6 and 7 so that digits do not repeat and the terminal digits are even, is
A
144
B
72
C
288
D
720
Answer :
720
Terminal digits are the first and last digits.
$$\therefore$$ Terminal digits are even
$$\therefore {1^{st}}$$ place can be filled in 3 ways and last place can be filled in 2 ways and remaining places can be filled in $$^5{P_4} = 120{\text{ ways}}{\text{.}}$$
Hence the number of six digit numbers so that
the terminal digits are even, is $$3 \times 120 \times 2 = 720$$
135.
Eight chairs are numbered 1 to 8. Two women and three men wish to occupy one chair each. First the women choose the chairs from amongst the chairs marked 1 to 4 ; and then the men select the chairs from amongst the remaining. The number of possible arrangements is
A
$$^6{C_3} \times {\,^4}{C_2}$$
B
$$^4{P_2} \times {\,^4}{C_3}$$
C
$$^4{C_2} + {\,^4}{P_3}$$
D
none of these
Answer :
none of these
$$\overline 1 \,\,\overline 2 \,\,\overline 3 \,\,\overline 4 \,\,\overline 5 \,\,\overline 6 \,\,\overline 7 \,\,\overline 8 $$
Two women can choose two chairs out of 1, 2, 3, 4, in $$^4{C_2}$$ ways and can arrange themselves in 2! ways. Three men can choose 3 chairs out of 6 remaining chairs in $$^6{C_3}$$ ways and can arrange themselves in 3! ways
∴ Total number of possible arrangements are
$$^4{C_2} \times 2!\,\, \times {\,^6}{C_3} \times \,3! = {\,^4}{P_2} \times {\,^6}{P_3}$$
136.
Total number of four digit odd numbers that can be formed using 0, 1, 2, 3, 5, 7 (using repetition allowed) are
A
216
B
375
C
400
D
720
Answer :
720
Required number of numbers
$$ = 5 \times 6 \times 6 \times 4 = 36 \times 20 = 720.$$
137.
The number of 4-digit numbers that can be made with the digits 1, 2, 3, 4 and 5 in which at least two digits are identical, is
A
$${4^5} - 5!$$
B
$$505$$
C
$$600$$
D
None of these
Answer :
$$505$$
The number of numbers when repetition is allowed $$ = {5^4}.$$
The number of numbers when digits cannot be repeated $$ = {\,^5}{P_5}.$$
∴ the required number of numbers $$ = {5^4} - 5!.$$
138.
From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be selected and arranged in a row on a shelf so that the dictionary is always in the middle. Then the number of such arrangement is :
A
at least 500 but less than 750
B
at least 750 but less than 1000
C
at least 1000
D
less than 500
Answer :
at least 1000
4 novels, out of 6 novels and 1 dictionary out of 3 can be selected in $$^6{C_4} \times {\,^3}{C_1}{\text{ ways}}{\text{.}}$$
Then 4 novels with one dictionary in the middle can be arranged in $$4!$$ ways.
$$\therefore $$ Total ways of arrangement
$$ = {\,^6}{C_4} \times {\,^3}{C_1} \times 4! = 1080.$$
139.
Let $$p$$ be a prime number such that $$p \geqslant 11.$$ Let $$n = p! + 1.$$ The number of primes in the list $$n + 1, n + 2, n + 3, ..... , n + p – 1,$$ is
A
$$p - 1$$
B
$$2$$
C
$$1$$
D
None of these
Answer :
None of these
For $$1 \leqslant i \leqslant p - 1,p!$$ is divisible by $$\left( {i + 1} \right).$$
Thus, $$n + i = p! + \left( {i + 1} \right)$$ is divisible by $$\left( {i + 1} \right)$$ for $$1 \leqslant i \leqslant p - 1$$
∴ None of $$n + 1, n + 2, ..... , n + p - 1$$ is prime.
140.
A bag contains 3 black, 4 white and 2 red balls, all the balls being different. The number of selections of at most 6 balls containing balls of all the colours is
A
$$42\left( {4!} \right)$$
B
$${{2^6} \times 4!}$$
C
$$\left( {{2^6} - 1} \right)\left( {4!} \right)$$
D
None of these
Answer :
$$42\left( {4!} \right)$$
The required number of selections
$$ = {\,^3}{C_1} \times {\,^4}{C_1} \times {\,^2}{C_1}\left( {^6{C_3} + {\,^6}{C_2} + {\,^6}{C_1} + {\,^6}{C_0}} \right).$$