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51.

Find the number of 4 digit odd numbers that can be formed using the digit 4,6,7,9,3 so that each digit occurs at most once in each number.A. 120B. 24C. 48D. 72

Answer» (i) The units digit of the number must be odd, i.e., it can be done in 3 ways.
(ii) Now find the number of ways in which the three digits can be filled using four digits.
(iii) Use the fundamental theorem of counting.
Discussion
52.

How many 3-digits numbers can be formed using the digits (2,4,5,7,8,9), if no digit occurs more than once in each number?A. 80B. 90C. 120D. 140

Answer» Here, we have 6 digit .
Since we have to find 3-digit numbers, the first digit can be filled in 6 ways, second digit can be filled in 5 and thrid digit can be filled in 4 ways.
(`because` No digit is repeated)
`therefore` The required number of digits
`=6xx5xx4=120`.
Discussion
53.

If a polygon has 6 sides, then the number of diagonals of the polygon is _______A. 18B. 12C. 9D. 15

Answer» The number of diagonals of a n-sided polygon is `(n(n-3))/(2)`.
Discussion
54.

How many two digit numbers can be formed using the digits (1,2,3,4,5) if no digit occurs more than once in each number?A. 10B. 20C. 9D. 16

Answer» r objects can be arranged out of n objects is `.^(n)p_(r)`. Ways.
Discussion
55.

If `.^(n)C_(4)=35`, then `.^(n)P_(4)=` ____A. 120B. 140C. 840D. 420

Answer» `.^(n)P_(r)=.^(n)C_(r)xxr!`
Discussion
56.

`.^(n)C_(3)`=___A. n!B. 1C. nnD. n

Answer» Use, `n^(n)C_(r)=(n!)/((n-r)!)`
Discussion
57.

If `.^(2n)C_(4): .^(n)C_(3)=21:1`, then find the value of n.A. 4B. 5C. 6D. 7

Answer» Given, `(.^(2n)C_(4))/(.^(n)C_(3))=(21)/(1)`
`.^(2n)C_(4)=21.^(n)C_(3)`
`((2n)!)/((2n-4)!4!)=21(n!)/((n-3)!3!)`.
`(2n)(2n-1)(2n-2)(2n-3)=84(n)(n-1)(n-2)`
`4(2n-1)(2n-3)=84(n-2)`.
Frm the options, n=5 satisfies the above equation.
Discussion
58.

From 8 gentlemen and 5 ladies, a committee of 4 is to be formed. In how many ways can this be done. (a) when the committee consists of exactly three gentlemen? (b) when the committee consists of at most three gentlemen?

Answer» (a) We have to select three out of 8 gentlemen and one out of 5 ladies. Hence, the number of ways in which this can be done `=.^(8)C_(3)xx.^(5)C_(1)=280`
(b) The committee is to contain at most three gentlemen, i.e., it may contain either 1,2 or 3 gentlemen.
Hence, the total number of ways `=.^(8)C_(1)xx.^(5)C_(3)+.^(8)C_(2)xx.^(5)C_(2)+.^(8)C_(3)xx.^(5)C_(1)=80+280+280=640`.
Discussion
59.

In how many ways can 3 vowels be selected from the letters of the word EQUACATION?A. 56B. 10C. 28D. 40

Answer» There are 5 vowvels and 3 are to be chosen.
Discussion
60.

In a class there are 20 boys and 25 girls. In how many ways can a pair of a boy and a girl be selected?A. 400B. 500C. 600D. 20

Answer» (i) The number of ways a boy and a girl can be selected individually is `.^(20)C_(r)` and `.^(25)C_(t)`.
(ii) Use the fundamenetal theorem of counting.
Discussion