IIT JEE

How many 4-letter words can be formed using a, b, c, d and e

(i) without repetition

(ii) with repetition.

(i) The number of words that can be formed is equal to the number of ways of filling the four places.

&⇒ 5´4´3´2=120
words can be formed when repetition is not allowed.

(ii) The number of words that can be formed is equal to the number of ways of filling the four places.

First place can be filled in 5 ways. If repetition is allowed, all the remaining places can be filled in 5 ways each.

&⇒ 5´5
´ 5 ´ 5 = 625 words can be formed when
repetition is allowed.

(a) In how many ways can we form 8-digit
numbers from the digits

0, 1, 2, 3, 4, 5, 6, 7? (Repetition not allowed)

(b) How many of these are even ?

(c) How many of these are divisible by 25?

(d) How many of these are less than 20000 ?

(e) Find the number of integers greater than 7000 that can be formed with the digits 3, 5, 7, 8 and 9, no digit is being repeated. .

(a) There are 8 places to be filled.

First place can not
be filled by 0. I.e. First place can be filled in 7 ways. 2^{nd} place
can be filled either by the 6 remaining digits or by 0 i.e. in 7 ways. 3^{rd}
place can be filled in 6 ways (by the 6 remaining digits) and so on.

7 ´ 7 ´ 6 ´ 5 ´ 4 ´ 3 ´ 2 ´ 1 = 7 ´ 7!

Total no. of possible permutations = 8! Of these, there are some permutations in which 0 comes in the first place i.e. they are actually 7 digit numbers. No. of such permutations = 1 ´ 7! (there is only 1 choice for the first place i.e. 0).

So that no. of 8-digit numbers = 8! - 7! = 7! (8 - 1) = 7 ´ 7!.

(b) Even nos. can be categorised into two categories.

(i) Last digit is 0. For the remaining 7 digits, there are 7 choices .

&⇒ 7! ways.

(ii) Last digit is
not 0. For the last place, there are 3 choices i.e. 2, 4 and 6. There are 6
choices for the 1^{st} place (excluding zero and the digit in last
place).

There are 6, 5, 4, 3, 2, and 1 choices for the middle positions (i.e. remaining digits).

So no. of such numbers = 6 ´ 6 ´ 5 ´ 4 ´ 3 ´ 2 ´ 1 ´ 3 = 18 ´ 6!.

So total no. of numbers = 7! + 18 ´ 6!.

(c) Last digits can only be 25, 50, 75 or 100. But two zeroes are not available 100 is rejected. .

(i) Numbers finishing with zero, last two digits can be 50 only. For the remaining 6 places, there are 6 choices. So there are 6! such numbers.

(ii) Number finishing with 5. Second last place can be filled by 2 or 7 only.

No. of choices for various places is

\ No. of numbers = 5 ´ 5! ´ 2.

Total number of numbers = 6! +10 ´ 5!

(d) Number of single-digit numbers = 8

Number of double-digit numbers = 7 ´ 7 = 49

Number of 3-digit numbers = 7´ 7 ´ 6 = 294

Number of 4-digit numbers = 7´ 7 ´ 6 ´ 5 = 1470

Number of 5-digit numbers = 1´ 7 ´ 6 ´ 5 ´ 4 = 840

Total number = 2661

(e) A five digit
integers is always greater than 7000. The number of such integer is ^{5}P_{5}
= 5! = 120. For a four digit integer to be greater than 7000, it must begin
with 7, 8 or 9. The number of such integer is (3) (^{4}P_{3} )
= 72. .

Hence, the required number of such integers is 120 + 72 = 192.

How many 3 letter words can be formed from the letters of the word CALCUTTA ?

CALCUTTA has letters CC, AA, TT, U,L. There are different number of cases where 3 letter words can be formed.

I. 3 alike NOT POSSIBLE.

II. 2 alike, 1 different.

III. 3 different.

For case (II), the
number of ways it can be done is ^{3}C_{1} ^{4}C_{1},
since there are 3 pairs of identical letters and 2 other different letters.
Now, these 3 letters can be arranged in 3!/2! ways since 2 of them are
identical. Hence, the total number for case (II) is.

^{3}C_{1}. ^{4}C_{1}.
(3!/2!) = 36. .

For case (III), the
number of ways we can choose 3 different letters from 5 different letters is ^{5}C_{3}.
These 3 letters can be arranged in 3! Ways, so that the number of ways = ^{5}C_{3}.
3!=60.

Hence the total number of letters that can be formed =36+60 = 96

In how may ways 3 boys and 15 girls can sit together in a row such that between any 2 boys at least 2 girls sit.

Let x_{1} girls sit before first
boy, x_{2} girls between first and second boy, x_{3} girls
between second and third boy and x_{4} after the third boy.

Now number of ways in which we can choose the places for boys is

same as the number of solution of the equation.

x_{1} + x_{2} + x_{3}
+ x_{4} = 15

Subject to condition x_{1} ≥ 0, x_{2} ≥ 2, x_{3} ≥ 2, x_{4} ≥ 0

\
Required number of ways = coefficient of x^{15} in

(1 + x + x^{2} + ) (x^{2} +
x^{3} + ...) (x^{2} + x^{3} + ..) (1 + x + x^{2}+...).

= coefficient of x^{11} in (1 - x)^{-4
}= ^{4 + 11 - 1}C_{4 - 1} = ^{14}C_{3}

Now 3 boys can be permuted in their places in 3! ways and 15 girls can be permuted in 15! ways.

So required number of ways = ^{14}C_{3}(3!)(15!)

In how may ways can
(2t + 1) identical balls be placed in 3 distinct boxes so that any two boxes
together will contain more balls

than the third.

The total number of ways to place the balls
disregarding the constraints is ^{2t+1+3-1}C_{3-1 }= ^{2t+3}C_{2
}ways.

The total number of ways to place the balls
so that the first box will have more balls than the other two is ^{t + 3 - 1}C_{3
- 1} = ^{t + 2}C_{2} ways.

(We place (t + 1) balls in the first box and then divide the rest of t balls in the 3 boxes arbitrarily).

The same result applies to the case of 2nd box holding more balls than 1st or 3rd combined and also for the 3rd box containing more balls than 1st and 2nd combined.

Hence required
number of ways = ^{2t + 3}C_{2} - 3^{ t + 2}C_{2}
= t/2(t + 1) .