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1

AIEEE 2003

If $${}^n{C_r}$$ denotes the number of combination of n things taken r at a time, then the expression $$\,{}^n{C_{r + 1}} + {}^n{C_{r - 1}} + 2\, \times \,{}^n{C_r}$$ equals
A
$$\,{}^{n + 1}{C_{r + 1}}$$
B
$${}^{n + 2}{C_r}$$
C
$${}^{n + 2}{C_{r + 1}}$$
D
$$\,{}^{n + 1}{C_r}$$

Explanation

Arrange it this way,

$$^n{C_{r + 1}} + 2.{}^n{C_r} + {}^n{C_{r - 1}}$$

$$= {}^n{C_{r + 1}} + {}^n{C_r} + {}^n{C_r} + {}^n{C_{r - 1}}$$

$$\left[ \, \right.$$ Now use the rule,

$$\left. {{}^n{C_r} + {}^n{C_{r - 1}} = {}^{n + 1}{C_r}} \right]$$

$$= {}^{n + 1}{C_{r + 1}} + {}^{n + 1}Cr$$

$$= {}^{n + 2}{C_{r + 1}}$$
2

AIEEE 2003

The number of ways in which 6 men and 5 women can dine at a round table if no two women are to sit together is given by
A
$$7!\, \times 5!\,\,$$
B
$$6!\, \times 5!$$
C
$$30!$$
D
$$5!\, \times 4!$$

Explanation

6 men can sit at the round table = $$\left( {6 - 1} \right)! = 5!$$ ways

Now at the round table among 6 men there are 6 empty places and 5 women can sit at those 6 empty positions.

So total no of ways 6 men and 5 women can dine at the round table

= $$5!\, \times {}^6{C_5} \times 5!$$

= $$5!\, \times 6 \times 5!$$

= $$5!\, \times 6!$$
3

AIEEE 2003

A student is to answer 10 out of 13 questions in an examination such that he must choose at least 4 from the first five questions. The number of choices available to him is
A
346
B
140
C
196
D
280

Explanation

Case 1 :

No of ways student can answer 10 questions = $${}^5{C_4} \times {}^8{C_6}$$ = 140

Case 2 :

No of ways student can answer 10 questions = $${}^5{C_5} \times {}^8{C_5}$$ = 56

$$\therefore$$ Total ways = 140 + 56 = 196
4

AIEEE 2002

The sum of integers from 1 to 100 that are divisible by 2 or 5 is
A
3000
B
3050
C
3600
D
3250

Explanation

According to this question, any number between 1 to 100 should be divisible by 2 or 5 but not by 2$$\times$$5 = 10.

Possible numbers between 1 to 100 divisible by 2 are 2, 4, 6, .... , 100

This is an A.P where first term = 2, last term = 100 and total terms = 50.

$$\therefore$$ Sum of the numbers divisible by 2

= $${{50} \over 2}\left[ {2 + 100} \right]$$

= 25$$\times$$102

= 2550

Possible numbers between 1 to 100 divisible by 5 are 5, 10, 15, .... , 100

$$\therefore$$ Sum of the numbers divisible by 5

= $${{20} \over 2}\left[ {5 + 100} \right]$$

= 10$$\times$$105

= 1050

And possible numbers between 1 to 100 divisible by 10 are 10, 20, 30, .... , 100

$$\therefore$$ Sum of the numbers divisible by 10

= $${{10} \over 2}\left[ {10 + 100} \right]$$

= 5$$\times$$110

= 550

$$\therefore$$ Required sum = 2550 + 1050 - 550 = 3050

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