Consider the following statements:I. The numb | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) For statement $I$: The number of positive integral solutions of $x_1+x_2+x_3+x_4=n$ is given by the formula $\binom{n-1}{r-1}$.Here, $n=10$ and $r

Vedclass · Accepted Answer

Only $II$ is true

Vedclass · Answer

(B) For statement $I$: The number of positive integral solutions of $x_1+x_2+x_3+x_4=n$ is given by the formula $\binom{n-1}{r-1}$.Here, $n=10$ and $r=4$, so the number of solutions is $\binom{10-1}{4-1} = \binom{9}{3} = \frac{9 	imes 8 	imes 7}{3 	imes 2 	imes 1} = 84$.Since $84 
eq 286$, statement $I$ is false.For statement $II$: The exponent of a prime $p$ in $m!$ is given by Legendre's formula $\sum_{k=1}^{\infty} [\frac{m}{p^k}]$.The exponent of $5$ in $25!$ is $[\frac{25}{5}] + [\frac{25}{25}] = 5 + 1 = 6$.The exponent of $2$ in $25!$ is $[\frac{25}{2}] + [\frac{25}{4}] + [\frac{25}{8}] + [\frac{25}{16}] = 12 + 6 + 3 + 1 = 22$.Since the exponent of $5$ is $6$ and the exponent of $2$ is $22$, the highest power of $10$ that divides $25!$ is $10^6$.Thus, $25! = 10^6 	imes k$, where $k$ is an integer not divisible by $10$.Therefore, statement $II$ is true.

Consider the following statements:
$I$. The number of positive integral solutions of $x_1+x_2+x_3+x_4=10$ is $286$.
$II$. If $25! = 10^n \times k, (k \in N)$, then $n=6$.
Which one of the following options is true?

Similar Questions

The number of seven-digit numbers that can be formed by using the digits $1, 2, 3, 5,$ and $7$ such that each digit is used at least once is:

The number of numbers greater than $5000$ and less than $9000$ that are divisible by $3$, which can be formed using the digits $0, 1, 2, 5, 9$ with repetition allowed, is . . . . . . .

The number of permutations of all the letters $AAAABBBC$ in which all the $A$'s appear together in a block of $4$ letters or all the $B$'s appear together in a block of $3$ letters,is-

If $x={ }^{16} C_5+{ }^{12} C_4, y=\sum_{r=1}^3{ }^{(20-r)} C_4, z=\sum_{k=1}^4{ }^{(16-k)} C_3$,then $x+y+z=$

The number of natural numbers lying between $1012$ and $23421$ that can be formed using the digits $2, 3, 4, 5, 6$ (repetition of digits is not allowed) and divisible by $55$ is $....$

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Consider the following statements:$I$. The number of positive integral solutions of $x_1+x_2+x_3+x_4=10$ is $286$.$II$. If $25! = 10^n \times k, (k \in N)$, then $n=6$.Which one of the following options is true?