(Each question 1 marks)

Question 11 Mark

Write the number of ways in which 5 boys and 3 girls can be seated in a row so that each girl is between 2 boys.

Answer

Since, the boys and girls are alternating. $\therefore$ 5 boys can be arranged in 5 places in (5! +5!) ways and 3 girls can be arranged in 3 places in (3! + 3!) ways Hence, the total number of ways = ( 5! + 5!) × (3! + 3!) = 2 × (5!) × 2 × (3!) = 4 × [5 × 4 × 3 × 2] × [3 ×2] = 4 × 120 × 6 = 4 × 720 = 2880

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Question 21 Mark

Write the number of words that can be formed out of the letters of the word 'COMMITTEE'.

Answer

There are 9 letters in the word 'COMMITTEE' out of which 2 are M's, 2 are T's 2 are E's and the rest are all distinct. $\therefore$ The required number of words $=\frac{9!}{2!\ 2!\ 2!}=\frac{9!}{(2!)^3}$ Hence, the total number of words are $\frac{9!}{(2!)^3}$

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Question 31 Mark

Write the number of ways in which 6 men and 5 women can dine at a round table if no two women sit together.

Answer

If all 11 persons are to be seated in a round table and in the round table of 11 positions there are exactly 5 places for women and 6 places for men. It is given that no two women sit together, ${ }^5 P _5$ this can be done in ways. The remaining 6 positions can be filled by the 6 men in ${ }^6 P _6$ ways. so, by the fundamental principle of counting, the number of seating arrangements as required is ${ }^5 P_5 \times{ }^6 P_6=5!\times 6!$ Hence, the number of ways in which 6 men and 5 worn en can dine at a round table are $5!\times 6!$

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Question 41 Mark

In how many ways 4 women draw water from 4 taps, if no tap remains unused?

Answer

Total number of wornen = 4 Total number of taps = 4 If no tap remains unused. $\therefore$ Number of ways = 4 × 3 × 2 × 1 = 4 Hence, 4 women draw water from 4 taps in 4! ways.

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Question 51 Mark

Write the number of numbers that can be formed using all for digits 1, 2, 3, 4.

Answer

Given digits are 1, 2,3,4 Number of numbers formed by all four digits $={ }^4 P _4=4 \times 3 \times 2 \times 1$
Required Numbars $=24$

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Question 61 Mark

Write the remainder obtained when 1! + 2! + 3! + ... + 200! is divided by 14.

Answer

First we will find the least factorial term divisible by 14. As 7!=7 × 6 × 5! is divisible by 14 leaving remainder zero. Hence terms 7! onwards can be written as multiple of 7!. 8!=8 × 7!, 9! = 9 × 8 × 7!... like ways 200! can also be written as multiple of 7!. So all the terms 7! onwards are divisible by 14 leaving remainder zero. = 1! + 2! + 3! + 4! + 5! + 6! = 1 + 2 + 6 + 24 + 120 + 720 = 873 Hence remainder obtained when 1! + 2! + 3! + ….+ 200! is divided by 14 is 5.

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Question 71 Mark

Write the number of ways in which 7 men and 7 women can sit on a round table such that no two women sit together.

Answer

Each of the seven men can be arranged amongst themselves in 7! ways.
The women can be arranged amongst themselves in seven places, in 6! ways (i.e. nothings can be arranged in (n - 1)! ways around a round table). By fundamental principle of counting, total number of ways = 7! × 6!

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Question 81 Mark

Write the number of arrangements of the letters of the word BANANA in which two N's come together.

Answer

There are 6 letters in the word 'BANANA' out of which 3 are A's 2 are N's and the rest are all distinct. Considering both N's together and treating them as one letter we have 5 letters. These 5 letters can be arranged in $\frac{5!}{3!}$ $=\frac{5\times4\times3!}{3!}$ $=20$Hence, the 5 letters can be arranged in 20 ways.

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Question 91 Mark

Write the total number of possible outcomes in a throw of 3 dice in which at least one of the dice shows an even number.

Answer

There dice are thrown. One dice has 6 possible outcomes including 3 even and 3 odd numbers. Number of outcomes in which at least are dice shows even number = (one dice shows even numbers) or ( 2 dice shows even number) or (There dice shows even numbers). = (3 × 6 × 6) + (3 × 3 × 6) + (3 × 3 × 3) = 108 + 54 + 27 = 189 Required Number = 189

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Question 101 Mark

In how many ways can 4 letters be posted in 5 letter boxes?

Answer

Total number of letters $=4$ Total number of letter boxes $=5 \therefore$ The number of ways in which 4 letters posted in 5 letters boxes $=5 \times 5 \times 5 \times 5=5^4$

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Question 111 Mark

Write the number of 5 digit numbers that can be formed using digits 0, 1 and 2.

Answer

Total number of digits $=3$ Now, zero can not be first digit of the 5 digit numbers. $\therefore$ The number of 5 digit numbers $=2 \times 3 \times 3 \times 3 \times 3=2 \times 3^4$

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Question 121 Mark

Write the number of all possible words that can be formed using the letters of the word 'MATHEMATICS'.

Answer

There are 11 letters in the word 'MATHEMATICS' out of which 2 are M's, 2 are A's, 2 are rs and the rest are all distinct. So, the required number of words $=\frac{11!}{2!\ 2!\ 2!}$ $=\frac{11!}{(2!)^3}$

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